In the figure (i) given below, the sides of the quadrilateral touch the circle. Prove that AB + CD = BC + DA.

Let p, Q, R, S be the points where the circle touches the sides of the quadrilateral as shown in the figure below:

From A, AP and AS are the tangents to the circle.

∴ AP = AS …(i) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From B, BP and BQ are the tangents to the circle.

∴ PB = BQ ….(ii) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From C, CR and CQ are the tangents to the circle.

∴ CR = CQ ….(iii) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From D, DR and DS are the tangents to the circle.

∴ DR = DS ….(iv) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

Adding L.H.S. and R.H.S. of equations (i), (ii), (iii) and (iv) we get,

⇒ AP + PB + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = BC + DA

Hence, proved that AB + CD = BC + DA.