NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 – Quadrilaterals

Class 9 Ganita Manjari Chapter 12 Quick Links:

• Think and Reflect
• Exercise Set 12.1
• Exercise Set 12.2
• Exercise Set 12.3
• Exercise Set 12.4
• End of Chapter Exercises
• Test Papers with Solutions (PDF)

• Class 9 Ganita Manjari Chapter 12 MCQs
• Download Class 9 Maths Offline App

Class 9 Maths Ganita Manjari Chapter 12 Think and Reflect Solutions

Answer:
Yes. Every ordinary quadrilateral, whether convex or non-convex, can tile the plane with congruent copies, without gaps or overlaps.
One construction is to rotate a copy through 180° about the midpoint of a side, so that it fits against the original along that side. Continue using half-turns about the midpoints of the exposed sides. This produces a repeating tiling.
The four different vertex angles can meet around a point because their sum is 360°. A non-planar four-sided figure is not a flat tile, and a crossed figure is not an ordinary quadrilateral tile.

Answer:
No. Not every figure shown is a quadrilateral in the usual sense.

• ABCD: Yes; it is a convex quadrilateral.
• NOPE: No; three named vertices lie on one straight side, so the outline is a triangle.
• SILY: No; some of the supposed sides overlap.
• DART: Yes; it is a non-convex quadrilateral, with a reflex internal angle at D.
• CUTS: It is a self-intersecting four-sided figure, excluded from the usual definition used in this chapter.
• OPENS: No; it is an open chain, not a closed four-sided figure.
• BENT: It is non-planar, so it is excluded from the usual planar definition.

A suitable definition is: A quadrilateral is a closed figure formed by joining four distinct points in one plane in order, with no three vertices collinear, such that adjacent sides meet only at their common endpoint and non-adjacent sides do not intersect.
This includes both convex quadrilaterals and non-convex quadrilaterals such as DART.

Answer:
Yes, but more conditions are needed than merely saying “four non-collinear points”.
Take four distinct points A, B, C, D in one plane, with no three collinear. Join AB, BC, CD and DA. Require adjacent sides to have only their shared endpoint in common and require opposite sides not to intersect.
These four segments then form quadrilateral ABCD. The conditions exclude overlapping sides, a crossed figure, an open chain and a non-planar figure, while allowing a non-convex quadrilateral.

Answer:
Let ∠A = ∠C = x and ∠B = ∠D = y.
Then 2x + 2y = 360°, so x + y = 180°.
Thus, each pair of adjacent angles is supplementary.

The converse is also true. If each pair of adjacent angles sums to 180°, then
∠A + ∠B = ∠B + ∠C, giving ∠A = ∠C;
∠B + ∠C = ∠C + ∠D, giving ∠B = ∠D.
Therefore the opposite angles are equal and Theorem 3 makes ABCD a parallelogram.
Completed statement: “If each pair of adjacent angles in a quadrilateral ABCD is supplementary, then ABCD is a parallelogram.”

Answer:
Let P, Q, R be the midpoints of AB, AC, BC respectively. The midpoint information gives AP = PB, AQ = QC and BR = RC, but it does not immediately give the lengths or directions of PQ, PR and QR. Thus we cannot yet apply a congruence test to all four small triangles using only these known equalities.
First prove the Midpoint Theorem. It gives PQ = BC/2, PR = AC/2 and QR = AB/2.
Now △APQ, △BPR, △CQR and △PQR each have side lengths AB/2, AC/2 and BC/2. They are congruent by SSS.
The useful idea is to establish one new segment property first, then use it to solve the larger problem.

Answer:
(1) If PQ ∥ BC and PQ = BC/2, then P and Q are the midpoints of AB and AC. A proof is given in End-of-Chapter Question 22(i).
(2) The line through the midpoint of one side, parallel to a second side, bisects the third side.
Proof: Let P be the midpoint of AB and let the parallel through P meet AC at Q. Let N be the actual midpoint of AC. The Midpoint Theorem gives PN ∥ BC. There is only one parallel to BC through P, so lines PN and PQ are the same. They meet AC at the same point; hence Q = N. Therefore AQ = QC.

(Hint: To arrange MC ∥ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ∥ CS?)
Answer:
Extend AM beyond M to S so that AM = MS. Thus M is the midpoint of AS.
In △ABS, P is the midpoint of AB and M is the midpoint of AS. By the Midpoint Theorem, PM ∥ BS. Since P, M, C are collinear, MC ∥ BS.
In △ACS, Q and M are midpoints of AC and AS. Therefore QM ∥ CS. Since Q, M, B are collinear, MB ∥ CS.
Hence BSCM is a parallelogram.

Let X be the intersection of BC and MS. Diagonals of a parallelogram bisect each other, so BX = XC and MX = XS.
Thus line AM meets BC at its midpoint X: AX is the third median, and it passes through M.
Also, AM = MS = 2MX, giving AM : MX = 2 : 1.
From △ABS, PM = BS/2 = MC/2, so CM : MP = 2 : 1.
From △ACS, QM = CS/2 = MB/2, so BM : MQ = 2 : 1.
Therefore the three medians are concurrent at M and each is divided in the ratio 2 : 1 from its vertex.

Answer:
The four internal angles of SOME add to 360°. Therefore, placing one copy of each angle around the common point fills one complete turn, with no angular gap or overlap near that point.
For a pattern that continues consistently, use half-turns about shared-side midpoints. These make neighbouring sides coincide exactly and place the vertex angles in a compatible order. Repeating the construction gives the tiling illustrated in the chapter.
Matching the angle sum at one point is a useful first check, but it does not alone prove that an arbitrary arrangement will extend to a whole-plane tiling.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.1 Solutions

(i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB.
(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A.
(iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.
Answer:
(i) Sides adjacent to AB: AD and BC. The side opposite to AB: CD.
(ii) Angles adjacent to ∠A: ∠B and ∠D. The angle opposite to ∠A: ∠C.
(iii) Opposite sides are two sides of a quadrilateral that have no common endpoint.
Opposite angles are two internal angles whose vertices are not joined by a side of the quadrilateral.

(Hint: use the opposite vertex as well.)
Answer:
The internal angle at a vertex is the angle between the two adjacent sides measured through the region enclosed by the quadrilateral near that vertex. It is the smaller angle at a convex vertex and the reflex angle, greater than 180°, at a non-convex vertex.

Here is a precise way to choose the correct angle at A using the other vertices, including the opposite vertex C:

• Draw triangle BCD and consider the smaller angle θ between rays AB and AD.
• If A lies inside triangle BCD, A is the inward-pointing vertex; define the internal angle at A to be 360° − θ.
• Otherwise, define the internal angle at A to be θ.
• Apply the same rule at each of the other vertices.

Do not simply choose the angle containing the ray towards the opposite vertex: in some non-convex quadrilaterals that rule gives the wrong angle.

Answer:
If AB ∥ DC: No. A simple quadrilateral with a pair of parallel opposite sides is convex. To see this, draw the two parallel lines. The connecting sides AD and BC must join their endpoints without crossing. The four vertices then occur around the boundary of a convex trapezium; none can be an inward-pointing vertex.

• If AB = CD: Yes. Equal opposite side lengths alone do not prevent a dent.
  Example: Plot A(0, 0), B(2, 0), C(1, 1), D(1, 3) and join them in that order. AB = CD = 2 units, but C lies inside triangle ABD, so ABCD is non-convex.
• If ∠A = ∠C: Yes. A symmetric dart can have equal convex angles at A and C and a reflex angle at B.
  Example: A(−2, 0), B(0, 1), C(2, 0), D(0, 3). Reflection in the vertical axis exchanges A and C and preserves the figure, so ∠A = ∠C; B is the inward-pointing vertex.

(Hint: the three lines divide the plane into 7 regions.)
Answer:
There are seven regions. The order of the vertices is fixed as A-B-C-D.

Thus, four regions give non-convex quadrilaterals, two give crossed quadrilaterals and one gives convex quadrilaterals.
To check each case, place D in the region, join A-B-C-D-A and examine whether opposite sides cross or one vertex lies inside the triangle of the other three. The type cannot change while D stays in the same region, because a change would require D to pass through one of the three boundary lines.

Answer:
No, if “self-intersecting” means an actual crossing of its straight sides.
Suppose opposite sides AB and CD meet. Two intersecting lines lie in one plane. Since A and B lie on the first line and C and D on the second, all four vertices lie in that plane. The same reasoning applies if AD and BC meet.
Therefore, an actually self-intersecting quadrilateral is planar. A drawing or projection of a non-planar quadrilateral can look crossed even when the spatial sides do not meet.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.2 Solutions

(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
Answer:
(i) True.
Adjacent angles of a parallelogram sum to 180° and opposite angles are equal. If one angle is 90°, all four are 90°, so it is a rectangle.

(ii) False.
Every rhombus has perpendicular diagonals, including rhombuses whose angles are 60° and 120°. Such a rhombus is not a square. An additional right angle or equality of its diagonals, would make it a square.

(iii) True.
In parallelogram ABCD, compare △ABC and △BAD. AB is common, BC = AD and AC = BD by the given condition. The triangles are congruent by SSS, so ∠ABC = ∠BAD. These adjacent angles also sum to 180°, hence both are 90°. Therefore ABCD is a rectangle.

Answer:
Given ∠BAC = ∠CAD.
Since AB ∥ CD, ∠BAC = ∠ACD.
Since AD ∥ BC, ∠CAD = ∠BCA.
Therefore ∠ACD = ∠BCA, so AC bisects ∠C.
Also, ∠BAC = ∠BCA in △ABC.

Hence AB = BC, because sides opposite equal angles are equal.
Opposite sides of a parallelogram are equal: AB = CD and BC = AD.
Combining these equalities gives AB = BC = CD = DA. Thus ABCD is a rhombus.

(i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
(ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus?
(iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle? (In the ancient Indian study of quadrilaterals, equality of diagonals was considered significant. Quadrilaterals were first classified according to whether their diagonals were equal or not, before considering equality of sides.)
Answer:
(i) Yes.
Here the condition means that AC bisects the angles at A and C and BD bisects the angles at B and D.
In △ABC and △ADC, the angles at A and C are equal in pairs and AC is common.
By ASA, the triangles are congruent. Thus AB = AD and BC = CD.
Similarly, comparing △ABD and △CBD using the angle bisectors at B and D gives AB = BC.
Therefore all four sides are equal, so ABCD is a rhombus.

(ii) Yes.
Diagonals that bisect each other make ABCD a parallelogram.
Let their intersection be O. Right triangles AOB and COB have AO = CO and common side OB.
They are congruent by SAS, so AB = BC. A parallelogram with equal adjacent sides is a rhombus.

(iii) No.
A non-rectangular isosceles trapezium has equal diagonals.
For example, ABCD gives equal diagonals AC = BD, but the angles are not all right angles.
Add the condition that ABCD is a parallelogram or equivalently add that its diagonals bisect each other. A parallelogram with equal diagonals is a rectangle, as proved in Question 1(iii).

Answer:
Using the intersections of the bisectors of adjacent angles, as in the figure.
In a parallelogram, ∠A + ∠B = 180°.
In the triangle formed by AB and the bisectors of ∠A and ∠B, the angle at their meeting point is
180° − (∠A/2 + ∠B/2) = 180° − 90° = 90°.

The same argument works for each pair of adjacent vertex-angle bisectors.
Opposite angles are equal and opposite sides are parallel, so the bisector lines of opposite angles are parallel.

Thus the four intersection points form a quadrilateral with four right angles: a rectangle.
If AB = BC, the original parallelogram is a rhombus. Its angle bisectors lie along its two diagonals and all four relevant intersections coincide at the diagonal intersection. There is then no rectangle with four distinct vertices. The condition AB ≠ BC excludes this collapsed case.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.3 Solutions

Answer:
(i) The Midpoint Theorem gives
PQ = BC/2 = BR = RC,
QR = AB/2 = AP = PB,
PR = AC/2 = AQ = QC.
Therefore, by SSS
△PQR ≅ △QPA ≅ △RBP ≅ △CRQ.

(ii) Yes.
Through P draw a line parallel to QR, through Q a line parallel to PR and through R a line parallel to PQ.
Call the intersections of the first and second lines A, the first and third B, and the second and third C.
For example, APQR is a parallelogram, so AP = RQ. Also PBRQ is a parallelogram, so PB = RQ. Thus AP = PB. Similarly AQ = QC and BR = RC.
Hence, the reconstructed triangle has P, Q, R as its side midpoints and is the required △ABC.

Answer:
Let MN meet AD at X. Since M and N are midpoints, MN ∥ BC by the Midpoint Theorem.

As D is on BC, MX ∥ BD. In △ABD, the line through midpoint M of AB parallel to BD bisects AD, by the converse of the Midpoint Theorem.
Thus AX = XD.
If D coincides with B or C, the conclusion follows directly from M or N being a midpoint.

Answer:
Let E be the midpoint of AD.
In △ADC, EG joins two midpoints, so EG ∥ DC and EG = DC/2.
In △ADB, EH joins two midpoints, so EH ∥ AB and EH = AB/2.

Since AB ∥ DC, EG and EH lie on the same line through E. Consequently E, G, H are collinear and GH ∥ AB.
In this trapezium, G and H lie on the same ray from E, giving GH = |CD − AB|/2.
If AB = CD, the equal-and-parallel-sides test makes ABCD a parallelogram. Its diagonals have the same midpoint, so G = H and GH is not a non-zero segment.
The assumption excludes that case.

(i) Show that PR and QS bisect each other.
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?
Answer:
(i) By Varignon’s theorem, PQRS is a parallelogram. Its diagonals PR and QS therefore bisect each other.
(ii) The Midpoint Theorem gives PQ = AC/2 and QR = BD/2.
If AC = BD, then PQ = QR.
Thus PQRS is a rhombus, whose diagonals are perpendicular: PR ⟂ QS.
The converse is also true.
If PR ⟂ QS, they are perpendicular diagonals of the parallelogram PQRS.
That makes PQRS a rhombus, so PQ = QR.
Therefore AC/2 = BD/2 or AC = BD.

(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. This will be relevant when we return to study tilings at the end of this chapter. (Hint: How will you place vertex A′? How will you place B′, C′ and D′?)
(ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral A′B′C′D′ you constructed is congruent to ABCD. You may need to show why S is collinear with the points A′ and D′ that you constructed and similarly for P, Q and R. (Hint: Use congruence of triangles, for example △SDR ≅ △SD′R.)
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.
Answer:
(i) Keep the original drawing available as a reference. On the copied PQRS, reproduce triangle SPA: use the lengths SP, PA and SA and place A′ on the corresponding side of SP, preserving the original orientation relative to Q and R.
Extend A′P beyond P to B′ with PB′ = PA′.
Extend B′Q beyond Q to C′ with QC′ = QB′.
Extend C′R beyond R to D′ with RD′ = RC′.
Join D′A′. This recreates the required quadrilateral.

(ii) The copied, oriented triangle SPA′ is congruent to the original triangle SPA. The same rigid placement that matches the copied parallelogram to the original therefore matches A′ to A.
Because P is a midpoint, B is uniquely the point obtained by a half-turn of A about P. Our construction makes B′ in exactly that way. Likewise, C′ and D′ are obtained by the corresponding half-turns about Q and R, so they match C and D.
Consequently A′B′C′D′ is congruent to ABCD. Since S was the midpoint of AD, it is also the midpoint of A′D′. Thus A′, S, D′ are collinear and SA′ = SD′. The same midpoint statements hold at P, Q and R by construction.

(iii) PQ ∥ AC, QR ∥ BD, PQ = AC/2 and QR = BD/2.
If PQRS is a square, PQ ⟂ QR and PQ = QR. Therefore AC ⟂ BD and AC = BD.
Conversely, if AC ⟂ BD and AC = BD, adjacent sides PQ and QR of the Varignon parallelogram are perpendicular and equal. Hence PQRS is a square.

Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.4 Solutions

(Hint: Read the first 3 sentences of ‘Think and Reflect’ in the section on tiling.) (There are many ways to tile the plane using a suitable irregular pentagon. The most recent method was found in 2015.)

Answer:
Each internal angle of a regular pentagon is (5 − 2) × 180°/5 = 108°.
At a point where tile corners meet, their angles must add to 360°.
But 360°/108° = 10/3 is not an integer. Three corners give 324°, leaving a 36° gap; four give 432°, causing an overlap.

Even allowing a corner to meet the interior of another tile’s edge does not help. That edge contributes 180°, leaving 180°, which is also not an integer multiple of 108°.

More generally, at a pentagon vertex we would need 108k + 180l = 360 with k a positive integer and l a non-negative integer. Dividing by 36 gives 3k + 5l = 10, which has no such solution.
Therefore congruent regular pentagons cannot tile the plane.

Answer:
One possible DART has D(1, 1), A(0, 3), R(0, 0) and T(3, 0), joined in the order D-A-R-T-D. D lies inside triangle ART, so it is the reflex vertex.
Make congruent copies. Place a half-turned copy across a side by rotating through 180° about that side’s midpoint. Repeat across exposed sides, keeping already placed copies in their existing positions. The copies fit into a repeating tiling; a finite portion is shown in following Figure.

Alternatively, use the Varignon parallelogram grid, place the equally oriented copies at the selected grid positions and fill the intervening gaps with half-turned copies.

Preference: I prefer the half-turn method because each new tile is located by a single rotation about a known midpoint.
The general justification, valid for this non-convex tile too.

Class 9 Maths Ganita Manjari Chapter 12 End-of-Chapter Exercises Solutions

(Hint: Can you use the parallelogram tiling in the introduction?)
Answer:
Take any triangle ABC. Rotate a congruent copy through 180° about the midpoint of BC. The two triangles together form a parallelogram, since the half-turn makes the appropriate opposite sides equal and parallel.
Tile the plane with translated copies of this parallelogram. Keep the diagonal separating its two constituent triangles in every copy. This gives a tiling entirely by copies of the original triangle.

Answer:
The two endpoints in Fig. 12.33 lie on ruled lines four equal spaces apart. Mark the point where the segment crosses the ruled line two spaces from either endpoint. This point is the midpoint.

Justification:
Draw a vertical through one endpoint and a horizontal through the other, forming a right triangle with the given segment as its sloping side. The middle ruled line passes through the midpoint of the vertical side and is parallel to the horizontal side. By the converse of the Midpoint Theorem, it also bisects the sloping side.

Answer:
AB and CD intersect at E.

(i) To show that ∠A + ∠B + ∠C + ∠D < 360°:
Let ∠AED = x. Since vertically opposite angles are equal, ∠BEC = ∠AED = x.
In △AED, using the angle-sum property of a triangle, ∠A + ∠D + x = 180°
Therefore, ∠A + ∠D = 180° − x. …(1)
Similarly, in △BEC, ∠B + ∠C + x = 180°
Therefore, ∠B + ∠C = 180° − x. …(2)
Adding (1) and (2), ∠A + ∠B + ∠C + ∠D = 360° − 2x.
Since x > 0°, we get
∠A + ∠B + ∠C + ∠D < 360°.
Hence proved.

(ii) Construction for an angle sum of 2°:
Yes, such a quadrilateral can be constructed.
1. Draw ∠AED = 179°.
2. Extend AE beyond E to a point B.
3. Extend DE beyond E to a point C.
4. Join AD and BC.
ABCD is now a self-intersecting quadrilateral, with AB and CD intersecting at E.
Using the result proved above,
∠A + ∠B + ∠C + ∠D
= 360° − 2 × 179°
= 360° − 358°
= 2°.
Thus, the required quadrilateral is possible.

Answer:
Given: ABCD is a parallelogram. P and Q lie on BD such that DP = BQ.
To prove: APCQ is a parallelogram.
Construction: Join AC and let it intersect BD at O.
Proof: Since the diagonals of a parallelogram bisect each other,
OA = OC and OD = OB.
Also, DP = BQ. (Given)
Subtracting equal lengths from OD and OB, we get
OD − DP = OB − BQ.
Therefore, OP = OQ.
Thus, OA = OC and OP = OQ.
Hence, AC and PQ, the diagonals of quadrilateral APCQ, bisect each other at O.
A quadrilateral whose diagonals bisect each other is a parallelogram.
Therefore, APCQ is a parallelogram. Hence proved.

Answer:
Let the crease meet AB at P and AC at Q.

Proof:
When the paper is folded so that A touches B, AP falls exactly on BP.
Therefore, AP = PB.
Hence, P is the midpoint of AB.
Also, ∠APQ and ∠BPQ coincide on folding. Therefore,
∠APQ = ∠BPQ.
Since A, P and B lie on a straight line,
∠APQ + ∠BPQ = 180°.
Thus, ∠APQ = ∠BPQ = 90°.
Therefore, PQ ⊥ AB.
Now, ∠ABC = 90°, so BC ⊥ AB.
Since PQ and BC are perpendicular to the same line AB,
PQ ∥ BC.
In △ABC, PQ passes through the midpoint P of AB and is parallel to BC.
Therefore, by the converse of the Midpoint Theorem, Q is the midpoint of AC.
Hence, AQ = QC.
Thus, the crease line gives the midpoints of both AB and AC. Hence proved.

(i) Draw medians AP, BQ and CR of △ABC, meeting in a common point M. Cut along each median to get 6 triangles. Show with justification how to assemble the 6 pieces into 3 congruent triangles! (Hint: Align △MPB and △MPC along equal sides PB and PC, matching P with itself and B with C.) Prove that you get a triangle. What are its side lengths?
(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show that each of the resulting 9 triangles has sides AB/3, BC/3 and AC/3.
The procedure and the result above were unnoticed till 2014, when they were discovered by Lee Sallows, an amateur mathematician!
(iii) Naturally, we can next ask about the possibility of cutting a triangle and reassembling it into 2 congruent triangles.
You were asked this question in the chapter Measuring Space: Perimeter and Area. Can you solve this now?
(Hint: Use median AP in △ABC and then a median of △APB.)
For any two given figures with straight sides, we can ask whether one figure can be cut (using only straight cuts) into pieces and reassembled to make the other. This notion is called scissors congruence. The figures need not be in the plane. Research on such questions at a higher level is ongoing.
Answer:
(i) Reassembling the pieces into 3 congruent triangles
Let P, Q and R be the midpoints of BC, CA and AB respectively.
Draw the medians AP, BQ and CR, meeting at M.
By the Centroid Theorem,
AM = 2MP, BM = 2MQ and CM = 2MR.

Cut along the three medians to obtain six triangles.
First, take △MPB and △MPC.
Since PB = PC, rotate △MPC through 180° about P so that C falls on B.
Let M move to X.
Then M, P and X lie on a straight line, and PX = PM.
Therefore, the two pieces together form △MBX.
Its side lengths are:
MB = BM,
BX = CM, because rotation does not change lengths,
MX = MP + PX = 2MP = AM.

Similarly:
• Pair △MQC with △MQA.
Rotate △MQA through 180° about Q, bringing A onto C.
Let M move to Y. This gives △MCY.

• Pair △MRA with △MRB.
Rotate △MRB through 180° about R, bringing B onto A.
Let M move to Z. This gives △MAZ.

For the second triangle,
MY = 2MQ = BM and CY = AM.
For the third triangle,
MZ = 2MR = CM and AZ = BM.
Thus, each assembled triangle has side lengths AM, BM and CM.
Therefore, the three triangles are congruent by SSS.
Their side lengths are:
2AP/3, 2BQ/3 and 2CR/3.

(ii) Repeating the procedure to obtain 9 triangles
Consider the joining segments BP, CQ and AR in Diagram 2.
Since P, Q and R are the midpoints of MX, MY and MZ respectively:
• BP is a median of △MBX, and BP = BC/2.
• CQ is a median of △MCY, and CQ = CA/2.
• AR is a median of △MAZ, and AR = AB/2.
These are medians to sides of lengths AM, BM and CM respectively.

Since the three assembled triangles are congruent, their corresponding medians are equal.
Hence, each assembled triangle has median lengths:
BC/2, CA/2 and AB/2.
Part (i) shows that the new triangles have side lengths equal to two-thirds of the medians of the triangle being cut.
Therefore, repeating the procedure gives triangles with sides:
(2/3) × (BC/2) = BC/3,
(2/3) × (CA/2) = CA/3,
(2/3) × (AB/2) = AB/3.
Each of the 3 triangles produces 3 more triangles.
Thus, we obtain 3 × 3 = 9 triangles, each having sides AB/3, BC/3 and AC/3. Hence proved.

(iii) Reassembling the triangle into 2 congruent triangles
Draw median AP of △ABC, where P is the midpoint of BC.
In △APB, draw median PR, where R is the midpoint of AB.
Cut along AP and PR. This gives three pieces:
△APC, △APR and △BPR.

Keep △APC aside.
Rotate △BPR through 180° about R so that B falls on A. Let P move to S.
Join this piece to △APR along AR.
Since P, R and S lie on a straight line, the two pieces form △APS.
Now, compare △APS and △PAC.
AP = PA. (Common side)
AS = BP = PC.
(Rotation preserves lengths, and P is the midpoint of BC.)
Also, P and R are the midpoints of BC and BA. By the Midpoint Theorem,
PR = AC/2.
Since RS = PR,
PS = PR + RS
= 2PR
= AC.
Thus, the three corresponding sides are equal.
Therefore, △APS ≅ △PAC by SSS.
Hence, the original triangle can be cut into three pieces and reassembled into 2 congruent triangles.

(i) If EF ∥ AB, then show that F is the midpoint of BC. Conclude that EF = (AB + CD)/2.
(ii) If F is the midpoint of BC, then show that EF ∥ AB.
There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
Answer:
(i) Let EF meet diagonal DB at M.

In △DAB, E is the midpoint of DA and EM ∥ AB. By the converse of the Midpoint Theorem, M is the midpoint of DB. Also EM = AB/2.
In △DBC, M is the midpoint of DB and MF ∥ DC. Hence F is the midpoint of BC, and MF = DC/2.
M lies between E and F, so EF = EM + MF = (AB + CD)/2.

(ii) Direct proof: Let M be the midpoint of BD.
In △DAB, EM ∥ AB by the Midpoint Theorem.
In △DBC, MF ∥ DC by the Midpoint Theorem.
Since AB ∥ DC, both EM and FM are the same line through M parallel to the bases. Thus E, M, F are collinear and EF ∥ AB.
Alternative: Draw the parallel to AB through E and call its intersection with BC F′. Part (i) shows that F′ is the midpoint of BC. Since the midpoint is unique, F′ = F; hence EF ∥ AB.
Either proof is valid. The direct proof is particularly useful because it also explains the formula for EF.

Answer:
AO = CO because diagonals of a parallelogram bisect each other.
Also ∠AOP = ∠COQ, as vertically opposite angles and ∠OAP = ∠OCQ because AB ∥ CD.
Thus △AOP ≅ △COQ by ASA,
So, OP = OQ [by CPCT].
As P, O, Q are collinear, O is the midpoint of PQ.

Other Proof: A half-turn about O maps the parallelogram to itself, taking AB to CD and P to Q. Therefore OP = OQ.

Answer:
Given: AD ∥ BC, AD = 3 cm and BC = 5 cm. E and F are the midpoints of AB and CD respectively.
Since EF joins the midpoints of the non-parallel sides, by the trapezium midpoint result,
EF ∥ AD ∥ BC, and
EF = (AD + BC)/2
= (3 + 5)/2
= 4 cm.
Also, EF lies halfway between AD and BC.
Therefore, trapeziums AEFD and EBCF have equal heights.
Let each height be h cm.
Using the formula for the area of a trapezium,
Area of AEFD = ½ × (AD + EF) × h
= ½ × (3 + 4) × h
= 7h/2 cm².
Area of EBCF = ½ × (EF + BC) × h
= ½ × (4 + 5) × h
= 9h/2 cm².
Therefore, Area of AEFD : Area of EBCF
= (7h/2) : (9h/2)
= 7 : 9.
Hence, the required ratio is 7 : 9.

(i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex?
(ii) How many of these quadrilaterals are self-intersecting? How many are convex?
Answer:
(i) The four points form 3 different quadrilaterals.
A quadrilateral remains the same if we change the starting vertex or trace its sides in the reverse order.
Starting at A, the six possible orders form these pairs:
• ABCD and ADCB represent the same quadrilateral.
• ABDC and ACDB represent the same quadrilateral.
• ACBD and ADBC represent the same quadrilateral.
Therefore, the number of different quadrilaterals is
6 ÷ 2 = 3.
These are ABCD, ABDC and ACBD.

(ii) There are two possible cases.
Case 1: No point lies inside the triangle formed by the other three points.

In the arrangement shown, ABCD follows the outside boundary and is convex.
In ABDC and ACBD, sides AC and BD cross each other.
Hence, both are self-intersecting.
Therefore:
Number of self-intersecting quadrilaterals = 2.
Number of convex quadrilaterals = 1.

Case 2: One point lies inside the triangle formed by the other three points.
Suppose D lies inside △ABC.

Each of the three quadrilaterals has a dent at D and its sides do not cross.
Therefore, all three are non-convex quadrilaterals.
Hence:
Number of self-intersecting quadrilaterals = 0.
Number of convex quadrilaterals = 0.
Number of non-convex quadrilaterals = 3.

(i) If PB = 2 find QC. (Hint: See the dotted line)
(ii) If PB = 1/3 find QC. (Hint: See the dotted lines)
(iii) If PB = 2/3 find QC.
(Hint: Divide both PB and AP suitably.)
(iv) Show that if AP/PB is a rational number, then AP/PB = AQ/QC. In Grade 10, you will prove this equality without assuming AP/PB is rational. That will require a new idea.
Answer:
For parts (i)–(iii), AP = 1 and AQ = √5. All the dotted lines below are drawn parallel to BC.

(i) When PB = 2
Mark the midpoint R of PB and draw RS ∥ BC, meeting AC at S.
Then, AP = PR = RB = 1.
In △ARS, P is the midpoint of AR and PQ ∥ RS.
By the converse of the Midpoint Theorem, AQ = QS.
Also, in trapezium PBCQ, R is the midpoint of PB and RS is parallel to its bases.
By the trapezium midpoint result, QS = SC.
Therefore, AQ = QS = SC = √5.
Hence, QC = QS + SC
= √5 + √5
= 2√5.

(ii) When PB = 1/3
Divide AP into three equal parts at R and S.
Draw RU and SV parallel to BC, meeting AC at U and V.
Then, AR = RS = SP = PB = 1/3.
In △ASV, R is the midpoint of AS and RU ∥ SV.
Therefore, AU = UV.
In trapezium RPQU, S is the midpoint of RP and SV is parallel to its bases.
Therefore, UV = VQ.
Similarly, in trapezium SBCV, P is the midpoint of SB and PQ is parallel to its bases.
Therefore, VQ = QC.
Thus, AU = UV = VQ = QC.
Now, AQ = AU + UV + VQ = 3QC.
Hence, QC = AQ/3 = √5/3.

(iii) When PB = 2/3
Divide AP into three equal parts at R and S.
Divide PB into two equal parts at T.
Draw RU, SV and TW parallel to BC.
Then, AR = RS = SP = PT = TB = 1/3.
Applying the same midpoint reasoning as in part (ii),
AU = UV = VQ = QW = WC.
Since AQ consists of three of these equal parts,
QW = WC = AQ/3 = √5/3.
Therefore, QC = QW + WC
= √5/3 + √5/3
= 2√5/3.

(iv) When AP/PB is rational
Let AP/PB = m/n, where m and n are positive integers.
Divide AP into m equal parts and PB into n equal parts.
These parts have the same length because AP/m = PB/n.
Through every division point, draw a line parallel to BC, meeting AC.
We now show that these lines divide AC into equal parts.
For any three consecutive division points on AB, the middle point is the midpoint of the other two. The corresponding parallel therefore bisects the interval on AC:
• For the first two intervals, this follows from the converse of the Midpoint Theorem.
• For each following pair of intervals, it follows from the trapezium midpoint result in Exercise 7.
Hence, all the corresponding parts on AC are equal.
Let the length of each such part be x.
Since P separates m parts from n parts on AB, Q separates m parts from n parts on AC.
Therefore, AQ = mx and QC = nx.
Thus, AQ/QC = mx/nx
= m/n
= AP/PB.
Hence, AP/PB = AQ/QC.

Answer:
(i) Let DM and BN meet AC at E and F respectively.

Since ABCD is a parallelogram,
AB = CD and AB ∥ CD.
M and N are the midpoints of AB and CD. Therefore,
MB = AB/2 = CD/2 = DN.
Also, MB ∥ DN.
Thus, quadrilateral MBND has one pair of opposite sides equal and parallel.
Hence, it is a parallelogram.
Therefore, DM ∥ BN.
In △ABF, M is the midpoint of AB and ME ∥ BF.
By the converse of the Midpoint Theorem, E is the midpoint of AF.
So, AE = EF. … (1)
Similarly, in △CDE, N is the midpoint of CD and NF ∥ DE.
Therefore, F is the midpoint of CE.
So, CF = FE. … (2)
From (1) and (2),
AE = EF = FC.
Hence, DM and BN trisect AC.

(ii) Three ways to trisect a given segment PQ:
Method 1: Using part (i)

Construction:

1. Choose a point R outside the line PQ. Join PR and RQ.
2. Through P, draw a line parallel to RQ. Through Q, draw a line parallel to PR. Let these lines meet at S.
3. Mark the midpoints M and N of PR and QS respectively.
4. Draw SM and RN, meeting PQ at E and F respectively.

Justification:
PRQS is a parallelogram and PQ is its diagonal.
Therefore, by part (i), PE = EF = FQ.
Hence, E and F trisect PQ.

Method 2: Using Exercise 11

Construction:

1. Draw a ray from P that does not lie along PQ.
2. Mark U, V and R successively on this ray such that PU = UV = VR.
3. Join RQ.
4. Through U and V, draw lines parallel to RQ, meeting PQ at E and F respectively.

Justification:
In △PRQ, UE ∥ RQ. By Exercise 11,
PE/EQ = PU/UR = 1/2.
Thus, EQ = 2PE, giving PQ = 3PE.
Therefore, PE = PQ/3.
Similarly, VF ∥ RQ gives
PF/FQ = PV/VR = 2/1.
Thus, PF = 2FQ, giving PQ = 3FQ.
Therefore, FQ = PQ/3.
Now, EF = PQ − PE − FQ
= PQ − PQ/3 − PQ/3
= PQ/3.
Hence, PE = EF = FQ.

Method 3: Using the Centroid Theorem

Construction:

1. Draw a line through Q different from PQ.
2. Choose points R and S on opposite sides of Q on this line such that QR = QS.
3. Join PR and PS to form △PRS.
4. Mark the midpoint T of PR. Draw ST, meeting PQ at G.
5. Mark the midpoint H of PG.

Justification:
Q is the midpoint of RS, so PQ is a median of △PRS. Similarly, ST is another median.
Therefore, their intersection G is the centroid.
By the Centroid Theorem, PG : GQ = 2 : 1.
Thus, PG = 2GQ.
Since H is the midpoint of PG,
PH = HG = PG/2 = GQ.
Hence, PH = HG = GQ, so H and G trisect PQ.

(i) Can you explain why this is so? See the geometric reasoning in the text.
(ii) There is one exception: In a very special case the Varignon parallelogram becomes a single segment. When will this happen?
(iii) Does your reasoning apply even when ABCD is non-planar?
Answer:
13. Yes, the theorem holds for both non-convex and self-intersecting quadrilaterals, except for the special case in part (ii).
(i) Explanation
Let P, Q, R and S be the midpoints of AB, BC, CD and DA respectively. Join AC and BD.

In △ABC, P and Q are the midpoints of AB and BC. By the Midpoint Theorem,
PQ ∥ AC and PQ = AC/2.
Similarly, in △ADC,
SR ∥ AC and SR = AC/2.
Therefore, PQ ∥ SR and PQ = SR.
Also, applying the Midpoint Theorem in △BCD and △BAD gives
QR ∥ BD and PS ∥ BD.
Hence, QR ∥ PS.
Thus, both pairs of opposite sides of PQRS are parallel. Therefore, PQRS is a parallelogram.
The same reasoning works whether ABCD is convex, non-convex or self-intersecting, because we apply the Midpoint Theorem to individual triangles.

(ii) When does the parallelogram become a single segment?
This happens when the diagonals AC and BD are parallel.

We know that
PQ ∥ AC and QR ∥ BD.
If AC ∥ BD, then the lines containing PQ and QR are both parallel to AC.
Since they pass through Q, they must be the same line.
Therefore, P, Q and R are collinear.
Similarly, PS ∥ BD, so S also lies on this line.
Hence, all four midpoints are collinear and the Varignon parallelogram becomes a single segment.
Conversely, if all four midpoints are collinear, PQ ∥ AC and QR ∥ BD imply AC ∥ BD.
Thus, this happens exactly when AC ∥ BD.

(iii) What if ABCD is non-planar?
Yes, the reasoning still applies.
Each of △ABC, △ADC, △BCD and △BAD lies in its own plane.
Therefore, we can apply the Midpoint Theorem to each triangle and obtain
PQ ∥ SR and QR ∥ PS.
The four midpoints also lie in one plane.
To see this, consider the plane through P, Q and R.
The line through P parallel to QR lies in this plane.
Since PS ∥ QR, S also lies in this plane.
Therefore, PQRS is a plane parallelogram even when ABCD is non-planar.

Answer:
No. P, Q, R and S need not be the midpoints of the respective sides.
Consider the following counterexample.
Let ABCD be a square of side 3 cm. Choose P, Q, R and S on AB, BC, CD and DA respectively such that
AP = BQ = CR = DS = 1 cm.
Then, PB = QC = RD = SA = 2 cm.

The four corner triangles △APS, △BQP, △CRQ and △DSR each have two sides of lengths 1 cm and 2 cm, with the included angle equal to 90°.
Therefore, these triangles are congruent by SAS.
Their corresponding third sides are equal, so
SP = PQ = QR = RS.
Thus, both pairs of opposite sides of PQRS are equal.
Hence, PQRS is a parallelogram.

However, each chosen point divides its side into lengths 1 cm and 2 cm.
Therefore, none of these points is a midpoint.
Hence, the proposed converse of Theorem 9 is false.

It would be good to be able to prove that △APQ ≅ △CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem (Theorem 7). Start by extending PQ up to a suitable point S.
Answer:
15. (i) Proof of the Midpoint Theorem
Given: P and Q are the midpoints of AB and AC respectively in △ABC.
To prove: PQ ∥ BC and PQ = BC/2.
Construction: Extend PQ beyond Q by a length equal to PQ.
Thus, QS = PQ. Join CS.
In △APQ and △CSQ,
AQ = CQ (Q is the midpoint of AC)
PQ = SQ (By construction)
∠AQP = ∠CQS (Vertically opposite angles)
Therefore, △APQ ≅ △CSQ by SAS.
Hence, AP = CS and ∠APQ = ∠CSQ.
The equal alternate interior angles give AP ∥ CS. Since A, P and B are collinear, PB ∥ CS.
Also, AP = PB because P is the midpoint of AB.
Thus, PB = CS.
Therefore, quadrilateral PBCS has one pair of opposite sides equal and parallel.
Hence, PBCS is a parallelogram. This explains why the suggested congruence is useful.
Its opposite sides are equal and parallel, so
PS ∥ BC and PS = BC.
Since P, Q and S are collinear, PQ ∥ BC.
Also, PS = PQ + QS = 2PQ.
Therefore, 2PQ = BC, giving PQ = BC/2.
Hence, the Midpoint Theorem is proved.

(ii) Proof of the converse of the Midpoint Theorem
Given: P is the midpoint of AB and the line through P parallel to BC meets AC at Q.
To prove: AQ = QC and PQ = BC/2.
Construction: Extend PQ beyond Q.
Through C, draw a line parallel to AB, meeting the extended line at S.
Now, PS ∥ BC and PB ∥ CS.
Therefore, PBCS is a parallelogram, giving
CS = PB and PS = BC.
Since AP = PB,
AP = CS.
In △APQ and △CSQ,
∠PAQ = ∠SCQ (Alternate interior angles, AP ∥ CS)
AP = CS
∠APQ = ∠CSQ (Alternate interior angles, AP ∥ CS)
Therefore, △APQ ≅ △CSQ by ASA.
Hence, AQ = CQ and PQ = SQ.
Thus, Q is the midpoint of AC.
Also, PS = PQ + QS = 2PQ.
Since PS = BC,
PQ = BC/2.
Hence, the converse of the Midpoint Theorem is proved.

Answer:
A converse reverses the condition and conclusion of a statement.

(Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?
Answer:
Join AC and treat it as a hinge between △ABC and △ADC.

During rotation, ∠B and ∠D remain unchanged.
∠A and ∠C decrease as the hinge closes and increase as it opens.
Since the triangles lie in different planes,
∠BAD < ∠BAC + ∠CAD
∠BCD < ∠BCA + ∠ACD.
Adding ∠ABC and ∠ADC, and using the angle sums of the two triangles, we get
∠A + ∠B + ∠C + ∠D < 180° + 180° = 360°.
To obtain an angle sum of 2°:
Take two congruent isosceles triangles ABC and ADC with common base AC, such that
∠ABC = ∠ADC = 0.5°.
Start with the triangles overlapping and open the hinge slightly until ∠BAD = 0.5°.
Now, AB = BC, AD = DC and BD is common. Therefore,
△BAD ≅ △BCD by SSS.
Hence, ∠BCD = ∠BAD = 0.5°.

Thus, the partially opened quadrilateral is non-planar and
∠A + ∠B + ∠C + ∠D
= 0.5° + 0.5° + 0.5° + 0.5°
= 2°.

Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in Fig. 12.42 so that the diagonals EO and E′O′ are collinear with O = E′. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EO′ until EO matches E′O′. Verify that your cutout exactly matches S′O′M′E′.

(i) Prove the exact match of SOME with S′O′M′E′ using four parallelograms.
(ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies in Fig. 12.42. These 9 copies meet each other only at vertices and once again the blank spaces among them also form 4-gons congruent to SOME!
Answer:
(i) Proof of the exact match:

The copies are congruent. Since EO and E′O′ have the same direction, equal corresponding angles show that corresponding sides are parallel.
Thus, each quadrilateral below has one pair of opposite sides equal and parallel:

Using their opposite sides, we get
SS′ = OO′ = MM′ = EE′ = EO.
These movements also have the same direction.
Therefore, sliding SOME by the distance EO in the direction E → O takes
S → S′, O → O′, M → M′ and E → E′.
Hence, the cutout exactly matches S′O′M′E′.

(ii) Construction of the tiling

1. Slide copies of SOME by the length EO in both directions along EO.
2. Slide two more copies by the length SM in both directions along SM.
3. From these four copies, make suitable slides along the other diagonal to place the four corner copies.

We now have nine coloured copies, meeting only at vertices.
Each of the four enclosed gaps matches a copy of SOME rotated through 180° about the midpoint of a side of the central copy. Therefore, all four gaps are congruent to SOME.

(i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is Yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c.
(Hint: Start by drawing a segment of length c.)
(ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side?
(iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
Answer:
(i) Construction of the triangle
Draw AB = c.
Draw arcs with centre A and radius b, and centre B and radius a.
If the arcs meet at C outside AB, join AC and BC.

Then AC = b, BC = a and AB = c.
Since a ≤ b ≤ c, a triangle exists exactly when a + b > c.
If a + b = c, the points lie on a straight line.
If a + b < c, the arcs do not meet.

(ii) Possible fourth-side lengths
In a quadrilateral, each side must be shorter than the sum of the other three.

Let the fourth side be x. We need
x < 2 + 5 + 11, so x < 18,
and
11 < 2 + 5 + x, so x > 4.
Therefore, the possible lengths are 4 < x < 18.

(iii) General rule
A quadrilateral exists exactly when its longest side is shorter than the sum of the other three sides.
Thus, arrange the lengths as a ≤ b ≤ c ≤ d and check whether a + b + c > d.
If this inequality holds, a quadrilateral can be formed. If it does not, no quadrilateral can be formed.

(i) How should we define a diagonal of an n-gon? How many diagonals does an n-gon have? Make a table for small values of n. A 3-gon has no diagonals. A 4-gon has 2. How many diagonals does a 5-gon have? A 6-gon? Try to find a pattern and guess the answers for n = 7 and n = 8. Check your guesses by systematic counting.
(ii) Can you guess a formula for the number of diagonals? How many diagonals get added when we increase the number of sides by 1? Can you now justify why the formula you guessed is true for all n?
Answer:
(i) A diagonal of an n-gon is a line segment joining two non-adjacent vertices.

Thus, a pentagon has 5 diagonals, a hexagon has 9, a heptagon has 14 and an octagon has 20.

(ii) Formula and justification
From each vertex, we cannot draw a diagonal to itself or to its two adjacent vertices.
Therefore, each vertex has n − 3 diagonals.
Counting from all n vertices gives n(n − 3).
However, each diagonal is counted twice, once from each endpoint.
Hence, Number of diagonals = n(n − 3)/2.
This reasoning works for every n ≥ 3.
When one side is added:
Insert a new vertex between two adjacent vertices A and B.
• The new vertex gives n − 2 diagonals.
• The old side AB becomes one additional diagonal.
Therefore, the number of diagonals added is (n − 2) + 1 = n − 1.
This explains the successive increases 2, 3, 4, 5, 6, … in the table.

Answer:
The sum of the interior angles of an n-gon is (n − 2) × 180°.

Proof:
Divide the polygon into triangles using diagonals that lie inside it and do not cross.

Suppose we draw d diagonals and obtain t triangles.
Each diagonal adds one region, so
t = d + 1.
Counting the sides of all the triangles, each polygon side is counted once and each diagonal twice.
Therefore,
3t = n + 2d
3t = n + 2(t − 1)
t = n − 2.
The angles of these triangles together form all the interior angles of the polygon.
Since each triangle has an angle sum of 180°,
Sum of interior angles = (n − 2) × 180°.
This works for both convex and non-convex polygons.

To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with △ABC and points P and Q on sides AB and AC respectively.
The theorem has two assumptions and two conclusions, which we have named for further discussion.
Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC.
Conclusions: (PRLL) PQ ∥ BC, and (HALF) PQ = BC/2.
Let us use these four named conditions to discuss various possible statements.
(i) A natural converse of the Midpoint Theorem would be: “If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true.” Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.)
(ii) To see Theorem 7 as a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: “Suppose P is the midpoint of side AB of △ABC and Q is a point on side AC. If Q is the midpoint of AC then PQ ∥ BC.” (Also PQ = BC/2, but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that Theorem 7 is what you get! (And PQ = BC/2 also follows.) Now write Theorem 7 in terms of the four named conditions.
(iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. “If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?”
Write this out in words. This is a precise version of question (1) stated before Theorem 7. Can you answer it?
In the text, why did we focus only on Theorem 7 among the multiple possibilities? Because Theorem 7 turns out to be particularly useful. It is still important to pursue various avenues while exploring a subject: often it is only after examining multiple possibilities that we can judge what is most valuable.
Answer:
22. (i) Statement:
If P and Q lie on AB and AC of △ABC, with PQ ∥ BC and PQ = BC/2, then P and Q are the midpoints of AB and AC.
Proof: Through C, draw CS ∥ AB, meeting PQ produced at S.

PBCS is a parallelogram. Therefore,
PS = BC = 2PQ and CS = PB.
Hence, PQ = QS.
Now, ∠APQ = ∠CSQ (alternate angles) and ∠AQP = ∠CQS (vertically opposite angles).
Thus, △APQ ≅ △CSQ by ASA.
Therefore,
AP = CS = PB and AQ = QC.
Hence, P and Q are the required midpoints.

(ii) Converse:
“Suppose P is the midpoint of AB and Q lies on AC. If PQ ∥ BC, then Q is the midpoint of AC.”
This is Theorem 7. It also gives PQ = BC/2.
In terms of the named conditions:
(P MID) and (PRLL) ⇒ (Q MID) and (HALF).

(iii) Statement:
“If P is the midpoint of AB and PQ = BC/2, must Q be the midpoint of AC, or must PQ ∥ BC?”
Neither conclusion follows in general.
Counterexample:
Take AC = 4 cm, AQ = 1 cm and AM = 2 cm.
Construct an equilateral triangle PQM. Extend AP to B so that AP = PB.

P and M are midpoints. Therefore, by the Midpoint Theorem,
PM = BC/2 and PM ∥ BC.
Since △PQM is equilateral,
PQ = PM = BC/2.
But AQ = 1 cm and QC = 3 cm, so Q is not the midpoint.
Also, PQ is a different line from PM through P, so PQ is not parallel to BC.
To check the combinations, apart from interchanging P and Q, the possible assumption pairs are:

• (P MID), (Q MID): original theorem.
• (PRLL), (HALF): part (i).
• (P MID), (PRLL): Theorem 7.
• (P MID), (HALF): the remaining case, examined above.

Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.
Answer:
We must prove that the copies cover the whole plane and their interiors do not overlap.
The following construction helps us prove all three methods.

Choose an internal diagonal EO of SOME. It divides the quadrilateral into △EOS and △EOM.
Rotate each triangle through 180° about the midpoint of OS and OM respectively.
The paired triangles form parallelograms EOXS and EOYM.
Repeat each parallelogram along EO in both directions.
This produces two strips without gaps or overlaps.
Add further strips alternately, aligning their equal boundary edges so that triangles across each boundary join to form SOME or its half-turned copy.
Continuing above and below covers the whole plane.
Since the parallelograms fill each strip exactly, joining their triangles into quadrilaterals preserves this complete coverage.

(i) First method: Half-turns about side midpoints
In this strip arrangement, the copy across each side is obtained by a 180° rotation about that side’s midpoint. Thus, the first method produces exactly this tiling.
Repeating across all sides reaches every strip, with no gaps or overlaps.

(ii) Second method: The Varignon parallelogram grid
By the Midpoint Theorem, each side of the Varignon parallelogram is parallel to, and half of, a diagonal of SOME.
Therefore, moving two grid steps shifts SOME by one full diagonal.
The selected copies occupy the same positions as in the strip arrangement.
The remaining spaces are precisely its half-turned copies. Hence this method also gives a complete tiling.

(iii) Third method: Sliding along the diagonals
Sliding SOME by either full diagonal places it at the same positions obtained in the second method. Repeating these slides in both directions and filling the spaces with half-turned copies therefore produces the same complete tiling.

Thus, all three methods cover the plane without gaps or overlaps.

Answer:
The shaded fraction is 1/5.

By the figure’s 90° rotational symmetry, PQRS is a square and the four surrounding triangles ABP, BCQ, CDR and DAS are congruent.
Since AE = CG and AE ∥ CG, AECG is a parallelogram.
Hence EQ ∥ AP.

In △ABP, E is the midpoint of AB.
Therefore, by the converse of the Midpoint Theorem, BQ = QP.
Also, congruent triangles ABP and BCQ give AP = BQ.

Let PQ = x. Then AP = x and BP = 2x.
Area of △ABP = ½ × AP × BP
= ½ × x × 2x = x².
The shaded square also has area x².
Thus, the whole square has area 5x².
Required fraction = x²/(5x²) = 1/5.

Download 10 Test Papers: If you have practice the chapter 12 thoroughly, not revise the chapter with the help of practice test papers with answer and solutions.