Question

prove that the parallogram cirumscribe a circle is rhombus ?​

Answer 1

Given :

➤ A circle with centre O

➤ A parallelogram ABCD touching the circle at point P, Q, R and S

To Prove :

➤ ABCD is a rhombus

Proof :

A rhombus is a parallelogram with all sides equal so, we have to prove all side equal.

In parallelogram ABCD

AB = CD and AD = BC ...❶ [opposite sides of parallelogram are equal]

From theorem :

Length of tangent drawn from external point are equal.

So ,

• AP = AS ...❷
• BP = BQ ...❸
• CR = CQ ...❹
• DR = DS ...❺

Adding equation ❷, ❸, ❹ and ❺ we get

⟾ AP + BP + CR + DR = AS + BQ + CQ + DS

⟾ (AP + BP) + (CR + DR) = +AS + DS) + (BQ + CQ)

⟾ AB + CD = AD + BC

⟾ CD = AB and BC = AB [from equation ❶]

⟾ AB + AB = AD + AD

⟾ 2AB = 2AD

⟾ AB = AD $\:$

So ,

⟾ AB = CD = AD = CD

Hence, ABCD is a rhombus.