Question

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

Answer 1

We know that the tangents drawn from an external point to a circle are equal.

$$\begin{gathered} \therefore AP=AS\dots \dots \left(i\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}A\right] \\ BP=BQ\dots \dots \dots .\left(ii\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}B\right] \\ CR=CQ\dots \dots \dots \dots .\left(iii\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}C\right] \\ DR=DS\dots \dots \dots ..\left(iv\right)\left[\mathrm{tangents}\mathrm{}\mathrm{from}D\right] \\ \therefore AB+CD=\left(AP+BP\right)+\left(CR+DR\right) \\ =\left(AS+BQ\right)+\left(CQ+DS\right)[\mathrm{using}\left(i\right),\left(ii\right),\left(iii\right)\mathrm{and}(iv\left)\right] \\ =\left(AS+DS\right)+\left(BQ+CQ\right) \\ =\left(AD+BC\right) \\ \mathrm{Hence},\left(AB+CD\right)=\left(AD+BC\right) \end{gathered}$$