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In figure, a quadrilateral $ABCD$ is drawn to circumscribe a circle, with center $O$, in such a way that the sides $AB,\ BC,\ CD$ and $DA$ touch the circle at the points $P,\ Q,\ R$ and $S$ respectively. Prove that $AB\ +\ CD\ =\ BC\ +\ DA$.
Given: A quadrilateral $ABCD$ is drawn to circumscribe a circle, with center O, in such a way that the sides $AB,\ BC,\ CD$ and $DA$ touch the circle at the points $P,\ Q,\ R$ and $ S$ respectively.
To do: To Prove that $AB\ +\ CD\ =\ BC\ +\ DA$.
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
$AP\ +\ BP\ +\ CR\ +\ DS\ =\ AS\ +\ BQ\ +\ CQ\ +\ DS$
$\therefore \ ( AP\ +\ BP) \ +\ ( CR\ +\ DR) \ =\ ( AS\ +\ DS) \ +\ ( BQ\ +\ CQ)$
$\therefore \ AB\ +\ CD\ =\ AD\ +\ BC$
$\therefore \ AB\ +\ CD\ =\ BC\ +\ DA\ \dotsc ..( proved)$
Updated on: 10-Oct-2022
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