Question

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC ?

Answer 1

Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure.

We know that, tangents drawn from an external point to the circle are equal in length.

Therefore,

`{:[AP,=,AS],[BP,=,BQ],[CQ,=,CR],[DR,=,DS]}`           ................(1)

∴AB + CD = (AP + BP) + (CR + DR)

= (AS + BQ) + (CQ + DS) [Using (1)]

= (AS + DS) + (BQ + CQ)

= AD + BC

Hence, AB + CD = AD + BC