Question

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

Answer 1

AG=AG(TANGENT FROM A)eq...1
GB=BI(FROM B)eq.....2
DJ=HD(FROM D)eq..3
JC=IC (FROM C)eq.....4
eq..1+2+3+4
AB+DC=DA+BC..

Answer 2

let the circle touch the side AB,BC,CD,DA of quadrilateral ABCD at point P,Q,R,S respectively.
AP=AS,DS=DR,CR=CQ,BQ=BP
and A-P-B,B-Q-C,C-R-D,A-S-D
now,AB+CD=AP+PB+CR+RD
=AS+BQ+CQ+DS
=AS+DS+BQ+CQ
=AD+BC
AB+CD=AD+BC