Question

A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig.) . Prove that $AB+CD=AD+BC$

Answer 1

Circle properties:

Let the$P,Q,R,S$be point of contacts for tangent $AB,BC,CD,DA$ respectively from the figure.

We know that the length of two tangents from one point is equal.

So,

$$\begin{gathered} AP=AS\dots \dots ..\left(1\right) \\ BP=BQ\dots \dots ..\left(2\right) \\ CR=CQ\dots \dots ..\left(3\right) \\ DR=DS\dots \dots ..\left(4\right) \end{gathered}$$

By, adding),$\left(1\right),\left(2\right),\left(3\right)$ and$\left(4\right)$ , we get

$$\begin{gathered} AP+BP+CR+DR=AS+BQ+CQ+DS \\ \Rightarrow (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)\mathbf{(}\mathbf{On}\mathbf{}\mathbf{Rearranging}\mathbf{)} \\ \Rightarrow AB+CD=AD+BC \end{gathered}$$

Hence, $\mathbf{AB}\mathbf{}\mathbf{+}\mathbf{}\mathbf{CD}\mathbf{}\mathbf{=}\mathbf{}\mathbf{AD}\mathbf{}\mathbf{+}\mathbf{}\mathbf{BC}$ proved.