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
Answers
Answered by
12
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
I think is this help u
Attachments:
Similar questions