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

Given: 

A quadrilateral $ABCD$ is drawn to circumscribe a circle

To do: 

We have to prove that $AB\ +\ CD\ =\ AD\ +\ BC$.

Solution:

Since tangents drawn from an exterior point to a circle are equal in length,

$AP\ =\ AS\ \dotsc .( 1)$

$BP\ =\ BQ\ \dotsc .( 2)$

$CR\ =\ CQ\ \dotsc .( 3)$

$DR\ =\ DS\ \dotsc .( 4)$

Adding equations $( 1) ,\ ( 2) ,\ ( 3)$ and $( 4)$, we get

$AP\ +\ BP\ +\ CR\ +\ DS\ =\ AS\ +\ BQ\ +\ CQ\ +\ DS$

$\therefore \ ( AP\ +\ BP) \ +\ ( CR\ +\ DR) \ =\ ( AS\ +\ DS) \ +\ ( BQ\ +\ CQ)$

$\therefore \ AB\ +\ CD\ =\ AD\ +\ BC$

Hence proved.

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Updated on: 10-Oct-2022

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