Question

If a circle touches all the four sides of a quadrilateral ABCD at
points P, Q, R, S. Then prove that AB+CD = BC+DA​

Answer 1

Step-by-step explanation:

ABCD Is a quadrilateral,

PQRS Is a points where touches circle

We know that two

tangents are equal,

DR=DS. ---1

CR=CQ. ----2

BP=BQ. ---3

AP=AS. ----4

ADD1,2,3 and4

DR+CR+BP+AP = DS+CQ+BQ+AS

(DR+CR)+(BP+AP)=(DS+AS)+(CQ+BQ)

CD+AB=AD+BC

AB+CD=AD+BC

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