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Question
A quadrilateral is drawn to circumscribe a circle. Prove that the sums of opposite sides are equal ?
Solution
Let ABCD be the quadrilateral circumscribing the circle.
Let E, F, G and H be the points of contact of the quadrilateral to the circle.
To Prove: AB + DC = AD + BC
Proof:
AB = AE + EB
AD = AH + HD
DC = DG + GC
BC = BF + FC
We have:
AE = AH (Tangents drawn from an external point to the circle are equal.)
Similarly, we have:
BE = BF
DH = DG
CG = CF
Now, we have:
AB + DC = AE + EB + DG + GC
= AH + BF + DH + CF
= (AH + DH) + (BF + CF)
= AD + BC
⇒ AB + DC = AD + BC
Thus, if a quadrilateral is drawn to circumscribe a circle, the sums of opposite sides are equal.
Hence, proved.
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