**Q) **A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC

**Ans: **

By tangents property, we know that the tangents drawn on a circle from an external point are always equal,

∴ from Point A: AP = AS ………….. (i)

from Point B: BP = BQ ………… (ii)

from Point C: CQ = CR ………… (iii)

from Point D: DR = DS ………… (iv)

Now let’s start from LHS: AB + CD

= (AP + PB) + (CR+ DR)

Substituting values from above 4 equal tangent values, we get:

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

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

= AD + BC

= RHS

**Hence Proved !**