Chapter 9.2, Problem 14WE

To determine

To prove: A quadrilateral ABCD is drawn to circumscribe a circle then prove that $AB+CD=AD+BC$ .

Explanation of Solution

Given information: Let ABCD be a quadrilateral circumscribing the circle with centre O. The quadrilateral touches the circle at point P, Q, R and S.

Formula used: Lengths of tangents drawn from external point are equal.

Proof: Consider the figure below,

  

According to theorem, lengths of tangents drawn from external point are equal.

Therefore,

  $\begin{array}{l}AP=AS\\ BP=BQ\\ CR=CQ\\ DR=DS\end{array}$

Adding above equations,

  $\begin{array}{l}AP+BP+CR+DR=AS+BQ+CQ+DS\\ \left(AP+BP\right)+\left(CR+DR\right)=\left(AS+SD\right)+CQ+BQ\\ AB+CD=AD+BC\end{array}$

Hence, it is proved.