In parallelogram ABCD, with AP and CQ as perpendiculars from vertices A and C on diagonal BD:
(i) Δ APB ≅ Δ CQD: In parallelogram ABCD, opposite angles are equal, so ∠ABD = ∠CDB. AP and CQ are perpendiculars, so ∠APB and ∠CQD are right angles. Since BD is a common side, triangles APB and CQD have a right angle, a congruent angle, and a shared side (hypotenuse). By the Right Angle-Hypotenuse-Angle (RHA) criterion, Δ APB is congruent to Δ CQD.
(ii) AP = CQ: From the congruence of Δ APB and Δ CQD, corresponding sides AP and CQ are equal, as they are the legs of these right-angled triangles.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that (i) ΔAPB ≅ ΔCQD

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Parallelogram ABCD and Perpendiculars AP and CQ

In the parallelogram ABCD, consider the perpendiculars AP and CQ drawn from vertices A and C, respectively, to the diagonal BD. This setup creates an interesting geometric scenario where the properties of parallelograms intersect with the principles of perpendicular lines and triangles. The perpendiculars introduce right angles into the structure, leading to the formation of right-angled triangles. These triangles, formed within the confines of the parallelogram, hold the key to uncovering deeper geometric relationships and congruences, which can reveal much about the nature and dimensions of ABCD.

Properties of Parallelogram ABCD

Parallelogram ABCD, by definition, has opposite sides that are equal and parallel, and opposite angles that are equal. These properties are fundamental to understanding the behavior of lines and angles within the parallelogram. When additional elements like perpendiculars from vertices to a diagonal are introduced, they interact with the parallelogram’s inherent properties to form new geometric shapes – in this case, right-angled triangles. The relationships between these triangles and the parallelogram’s sides and angles are crucial in exploring the congruences and equalities that arise.

Constructing Right-Angled Triangles APB and CQD

The perpendiculars AP and CQ from vertices A and C to diagonal BD create two right-angled triangles, APB and CQD, within parallelogram ABCD. In these triangles, AP and CQ are the perpendicular heights, and BD is the hypotenuse for both. The right angles at P and Q introduce a degree of symmetry and congruence between these triangles. This symmetry is not immediately apparent but becomes evident when considering the properties of the parallelogram and the specific nature of these right-angled triangles.

Proving Δ APB ≅ Δ CQD
To prove that triangles APB and CQD are congruent, we observe that in parallelogram ABCD, ∠ABD equals ∠CDB, as they are opposite angles. Since AP and CQ are perpendiculars to BD, ∠APB and ∠CQD are right angles. With BD as a common side (the hypotenuse of both triangles), triangles APB and CQD have a right angle, a congruent angle, and a shared side. According to the Right Angle-Hypotenuse-Angle (RHA) criterion for congruence, this is sufficient to establish that Δ APB is congruent to Δ CQD.

Establishing AP Equals CQ

From the congruence of triangles APB and CQD, it follows that corresponding parts of congruent triangles are equal. Therefore, the length of AP in triangle APB is equal to the length of CQ in triangle CQD. This equality is significant as it not only demonstrates a symmetrical relationship within the parallelogram but also reveals that the perpendicular distances from opposite vertices to the diagonal are the same. This finding is a direct consequence of the specific geometric construction within parallelogram ABCD and the properties of right-angled triangles.

Geometric Relationships in Parallelogram ABCD

In conclusion, the introduction of perpendiculars AP and CQ from vertices A and C to diagonal BD in parallelogram ABCD leads to the formation of two congruent right-angled triangles, APB and CQD. The congruence of these triangles, established through geometric principles, demonstrates that AP is equal to CQ. This equality and congruence within ABCD highlight the intricate geometric relationships that can arise in seemingly simple shapes like parallelograms. It showcases the depth of geometric principles and their ability to reveal hidden symmetries and properties in geometric figures.

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Questions of 9th Maths Exercise 8.1 in Detail

If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Show that the diagonals of a square are equal and bisect each other at right angles.
Diagonal AC of a parallelogram ABCD bisects ∠A. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that: (i) Δ APD ≅ Δ CQB (ii) AP = CQ (iii) Δ AQB ≅ Δ CPD (iv) AQ = CP (v) APCQ is a parallelogram.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ.
ABCD is a trapezium in which AB || CD and AD = BC. Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD.