Question

If a circle touches all the four sides of a quadrilateral ABCD at the points P,Q,R,S then prove that AB+CD= BC+AD.
draw a rough diagram

Answer 1

GIVEN_ ABCD is quadrilateral
circle with radius r is circumscribed in it
TO PROVE_AB+CD=BC+AD
CONST._join radius to point of contact at P,Q,R AND S

PROOF_
we can prove this by tangent theorem
AP=AS (tangents from the same point are equal)...1
similarly
PB=BQ...2
DR=DS...3
RC=CQ...4

adding 1,2,3 and 4
we have,

AP+PB+DR+RC=AS+SD+BQ+QC
(AP+PB)+(DR+RC)=(AS+SD)+(BQ+QC)
AB+DC=AD+BC


sorry I forgot to attach diagram