Prove that the parallelogram circumscribing a circle is a rhombus.

[IMAGE_1]

Question

Prove that the parallelogram circumscribing a circle is a rhombus.

[IMAGE_1]

Accepted Answer

Step 1: We know that lengths of tangents drawn from an external point to a circle are equal.
$AP = AS$ ... (1)
$BP = BQ$ ... (2)
$CR = CQ$ ... (3)
$DR = DS$ ... (4)
Step 2: Adding (1), (2), (3) and (4), we have
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
$\Rightarrow AB + CD = BC + AD$
Step 3: Since $ABCD$ is a parallelogram, $AB = CD$ and $BC = AD$.
$	herefore AB = CD$ and $BC = AD$
Step 4: $	herefore AB + AB = BC + BC \Rightarrow 2AB = 2BC \Rightarrow AB = BC$
$	herefore AB = BC = CD = AD$. Hence, $ABCD$ is a rhombus.

Question

Solution