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Question
If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair.
Solution
Consider a quadrilateral ABCD touching circle with center O at points E, F, G and H as in figure.
We know that
The tangents drawn from same external points to the circle are equal in length.
1. Consider tangents from point A [AM ⊥ AE]
AH = AE …. (i)
2. From point B [EB & BF]
BF = EB …. (ii)
3. From point C [CF & GC]
FC = CG …. (iii)
4. From point D [DG & DH]
DH = DG …. (iv)
Adding (i), (ii), (iii), & (iv)
(AH + BF + FC + DH) = [(AC + CB) + (CG + DG)]
⇒ (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG)
⇒ AD + BC = AB + DC [from fig.]
Sum of one pair of opposite sides is equal to other.
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