Question

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

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Solution

## 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,$AP=AS\dots \dots ..\left(1\right)\phantom{\rule{0ex}{0ex}}BP=BQ\dots \dots ..\left(2\right)\phantom{\rule{0ex}{0ex}}CR=CQ\dots \dots ..\left(3\right)\phantom{\rule{0ex}{0ex}}DR=DS\dots \dots ..\left(4\right)$By, adding),$\left(1\right),\left(2\right),\left(3\right)$ and$\left(4\right)$ , we get$AP+BP+CR+DR=AS+BQ+CQ+DS\phantom{\rule{0ex}{0ex}}⇒\left(AP+BP\right)+\left(CR+DR\right)=\left(AS+DS\right)+\left(BQ+CQ\right)\mathbf{\left(}\mathbf{On}\mathbf{}\mathbf{Rearranging}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}⇒AB+CD=AD+BC$Hence, $\mathbf{AB}\mathbf{}\mathbf{+}\mathbf{}\mathbf{CD}\mathbf{}\mathbf{=}\mathbf{}\mathbf{AD}\mathbf{}\mathbf{+}\mathbf{}\mathbf{BC}$ proved.

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