Question

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

Answer 1

Answer:

It is true

Step-by-step explanation:

The figure shows that the tangents drawn from the exterior point to a circle are equal in length.

As DR and DS are tangents from exterior point D so, DR = DS---- (1)

As AP and AS are tangents from exterior point A so, AP = AS---- (2)

As BP and BQ are tangents from exterior point B so, BP = BQ---- (3)

As CR and CQ are tangents from exterior point C so, CR = CQ---- (4)

Adding the equation 1,2,3 & 4, we get

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

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

CD+AB=DA+BC

AB+CD=BC+DA