Mathematics — Standard (041) — CBSE Class 10 board question

Given: Parallelogram ABCD circumscribes a circle.

To prove: ABCD is a rhombus, i.e., AB = BC = CD = DA.

Proof:

Since tangents drawn from an external point to a circle are equal in length:

From A: AP = AS
From B: BP = BQ
From C: CQ = CR
From D: DR = DS

Adding all:

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

$$AB + CD = AD + BC \quad \dots(1)$$

Since ABCD is a parallelogram:

$$AB = CD \quad \text{and} \quad AD = BC \quad \dots(2)$$

From (1) and (2):

$$2AB = 2AD$$

$$\Rightarrow AB = AD$$

Since AB = CD and AD = BC and AB = AD, all four sides are equal.

∴ ABCD is a rhombus. [Hence proved]

Source: Chapter 10, Section 10.3 (Theorem 10.2)

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• The key property used is Theorem 10.2: tangents from an external point are equal. Apply it at all four vertices.
• Adding the four pairs gives AB + CD = AD + BC (same as Q.8 in Exercise 10.2).
• Then substitute the parallelogram property (opposite sides equal) to get AB = AD, making all sides equal → rhombus.
• Label the points of contact clearly (P on AB, Q on BC, R on CD, S on DA) — examiners expect this diagram/labelling step.