Question
Prove that parallelogram circumscribing a circle is a rhombus.
✓
Answer
Given $ABCD$ is a parallelogram in which all the sides touch a given circle
To prove:- $ABCD$ is a rhombus
Proof:-
$\because$ $ABCD$ is a parallelogram
$\therefore AB = DC$ and $AD = BC$
Again $AP, AQ$ are tangents to the circle from the point A
$\therefore AP = AQ$
Similarly, $BR = BQ$
$CR = CS$
$DP = DS$
$\therefore (AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS)$
$\Rightarrow AD + BC = AB + DC$
$\Rightarrow BC + BC = AB + AB [\because AB = DC, AD = BC]$
$\Rightarrow 2BC = 2AB$
$\Rightarrow BC = AB$
Hence, parallelogram $ABCD$ is a rhombus
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