# Prove that the parallelogram circumscribing a circle is a rhombus

**Solution:**

ABCD is a parallelogram. Therefore, opposite sides are equal.

AB = CD

BC = AD

According to Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Therefore,

BP = BQ (Tangents from point B)…… (1)

CR = CQ (Tangents from point C)…… (2)

DR = DS (Tangents from point D)…… (3)

AP = AS (Tangents from point A)……. (4)

Adding (1) + (2) + (3) + (4)

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

On re-grouping,

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

AB + CD = BC + AD

Substitute CD = AB and AD = BC since ABCD is a parallelogram, then

AB + AB = BC + BC

2AB = 2BC

AB = BC

∴ AB = BC = CD = DA

This implies that all the four sides are equal.

Therefore, the parallelogram circumscribing a circle is a rhombus.

**Video Solution:**

## Prove that the parallelogram circumscribing a circle is a rhombus

### NCERT Solutions Class 10 Maths - Chapter 10 Exercise 10.2 Question 11:

Prove that the parallelogram circumscribing a circle is a rhombus

It has been proved that the parallelogram ABCD circumscribing a circle with center O is a rhombus