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Question

A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC  [1 MARK]

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Solution

Concept: 1 Mark

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

Therefore,  AP = AS                     ....(i)  [tangents from A]

BP = BQ                     ....(ii)  [tangents from B]

CR = CQ                     ....(iii) [tangents from C] 

DR = DS                     ....(iv) [tangents from D]

Therefore,    AB + CD = (AP + BP) + (CR + DR)

= (AS + BQ) + (CQ + DS)

[using (i), (ii), (iii), (iv)]

= (AS + DS) + (BQ + CQ)

= (AD + BC).

Hence, (AB + CD) = (AD + BC)

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