Question
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC
✓
Answer
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
AP = AS, ... (i) [tangents from A]
BP = BQ, ... (ii) [tangents from B]
CR = CQ, ... (iii) [tangents from C]
DR = DS. ... (iv) [tangents from D]
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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$B_5$
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$B_6$ |
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$B_7$
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5
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