To prove that in a quadrilateral ABCD, which circumscribes a circle, AB + CD = AD + BC:
Tangency Points: Let the points where the circle touches the quadrilateral be P, Q, R, and S, starting from the side AB and moving clockwise.
Equal Tangents: From each vertex, the tangents to a point of contact with the circle are equal in length. Therefore, AP = AS, BQ = BP, CR = CQ, and DS = DR.
Sum of Sides: AB = AP + BP, BC = BQ + CQ, CD = CR + DR, AD = AS + DS.
Substituting Equal Lengths: Substitute the equal tangents into the sum of sides: AB = AS + BP, BC = BP + CR, CD = CR + DR, AD = AS + DR.
Adding Equal Sides: Add AB and CD, and AD and BC separately: AB + CD = (AS + BP) + (CR + DR), AD + BC = (AS + DR) + (BP + CR).
Result: Since the sums are equal, AB + CD = AD + BC.
Thus, in a quadrilateral circumscribing a circle, the sum of the lengths of one pair of opposite sides is equal to the sum of the lengths of the other pair of opposite sides.

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Let’s discuss in detail

Introduction to Circumscribed Quadrilaterals

In the study of geometry, a fascinating concept is that of a quadrilateral circumscribing a circle. This occurs when a circle fits perfectly inside a quadrilateral, touching all four sides. An intriguing property of such a configuration is that the sum of the lengths of one pair of opposite sides of the quadrilateral is equal to the sum of the lengths of the other pair of opposite sides. This property is not immediately apparent and requires a deeper understanding of the properties of tangents and circumscribed figures.

Understanding Tangents in Circumscribed Figures

A key aspect of circumscribed figures is the role of tangents. In a quadrilateral ABCD circumscribing a circle, each side of the quadrilateral acts as a tangent to the circle at a specific point. Tangents from a common external point to a circle are equal in length. This property of tangents is crucial for proving the relationship between the sides of the quadrilateral.

Identifying Tangency Points

Consider the quadrilateral ABCD with the circle touching its sides at points P, Q, R, and S. These points are located on sides AB, BC, CD, and DA, respectively. The tangency points create segments on the sides of the quadrilateral, which are essential in our proof. For instance, AP and AS are segments on side AD, created by the tangency point S.

Applying the Property of Tangents
According to the property of tangents, the lengths of tangents drawn from an external point to a circle are equal. Therefore, in our quadrilateral, AP = AS, BQ = BP, CR = CQ, and DS = DR. These equalities are fundamental to establishing the relationship between the sides of the quadrilateral.

Summing the Sides of the Quadrilateral

To utilize these equalities, we express each side of the quadrilateral in terms of the segments formed by the tangency points. For example, side AB can be expressed as AP + PB, and side AD as AS + SD. Doing this for all sides and substituting the equal lengths, we find that the sum of the lengths of AB and CD is equal to the sum of the lengths of AD and BC.

Proving the Quadrilateral Property

By adding the lengths of opposite sides and using the property of tangents, we conclude that AB + CD = AD + BC in the circumscribed quadrilateral ABCD. This elegant result showcases the harmony and interconnectedness of geometric principles. It highlights how properties of tangents and circumscribed figures come together to reveal surprising and non-intuitive truths about geometric shapes.

Discuss this question in detail or visit to Class 10 Maths Chapter 10 for all questions.
Questions of 10th Maths Exercise 10.2 in Detail

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
If TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90°
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50° (B) 60° (C) 70° (D) 80°
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
If XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Prove that the parallelogram circumscribing a circle is a rhombus.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.