Question

In the given figure, if two chords PQ and RS of a circle with centre O intersect each other at M such that PM = MS, then prove that MR = MQ.

Answer 1

Given:

The two chords PQ and RS of a circle with centre O intersect each other at M such that PM = MS

To prove:

MR = MQ

Construction:

Join PR and SQ

Solution:

After joining the lines you will get the triangles named as:

ΔPMR and ΔSMQ

Here we will prove further by the concept of the congruent triangles.

So let's have ΔPMR and ΔSMQ

• PM = SM [Given]
• ∠SMQ = ∠PMR [Vertically opposite angles]

As we know that the angles subtended by the chord on any point on the boundary of the circle are same.

so,

• ∠MSQ = ∠MPR [For the chord QR]

ΔPMR ≅ ΔSMQ [By ASA]

∴ MR = MQ [By CPCT]

Hence proved.

Answer 2

Answer:

For △PMS and △SMQ

1.∠PMR=∠QMS (Vertically opposite angle)

2.∠MPR=∠MSQ (Angle subtended by the same arc are equal)

3.∠MRP=∠MQS (Sum property of △).

Thus By AAA criteria △PMS ≈ △SMQ.

ANS- Option C