Question

A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA

Answer 1

Given, a quad. ABCD and a circle touches its all four sides at P,Q, R, and S respectively.



To prove: AB + CD = BC + DA
Proof:
We know that tangents drawn from external point to a circle are equal.
So, AP=AS,
BP=BQ,
CQ=CR,
DR=DS.

L.H.S. = AB + CD

= AP + PB + CR + RD

= AS + BQ + CQ + DS

(Tangents from same external point are always equal)

= (AS + SD) + (BQ + QC)

= AD + BC

= R.H.S.
Hence Proved.