Question

If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair.

Answer 1

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.