Question

If the sides of a quadrilateral ABCD touch a circle, prove that:

AB + CD = BC + AD.

If the sides of a quadrilateral touch a circle as shown in figure, show that:

AB + CD = AD + BС

Answer 1

Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

Since AP and AS are tangents to the circle from the external point A

AP = AS  ...(i)

Similarly, we can prove that:

BP = BQ  ...(ii)

CR = CQ  ...(iii)

DR = DS   ...(iv)

Adding,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC