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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]
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)
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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