Prove that a Parallelogram Circumscribing a Circle is a Rhombus

Given:

• Parallelogram ABCD
• Circle PQRS

To prove: ABCD is a rhombus.

Proof:

We know that the tangents drawn from an exterior point to a circle are equal to each other. Therefore:
AP = AS ⇢ (1)
BP = BQ ⇢ (2)
DS = DR ⇢ (3)
CR = CQ ⇢ (4)

Adding the LHS and RHS of equations 1, 2, 3, and 4:
AP + BP + DS + CR = AS + BQ + DR + CQ
AB + DR + CR = AS + DS + BC
AB + CD = AD + BC

Since the opposite angles of a parallelogram are equal:
2AB = 2BC
AB = BC, and similarly, CD = AD.

Therefore: AB = CD = BC = AD.

Since all the sides are equal, ABCD is a rhombus.

Read More:

• Parallelogram
• Circle
• Tangent