Question

In a circle with centre O, chord SR = chord SM. The radius OS intersects the chord RM at P. Prove that PR = PM.

Answer 1

Given: In a circle with centre O, chord SR = chord SM. The radius OS meets the chord RM at P.

To Prove: PR = PM.

Proof (Step-wise):

1. SR = SM   ...(Given)

2. Equal chords subtend equal angles at the centre, so ∠ROS = ∠SOM.

3. Therefore, OS bisects ∠ROM; i.e. ∠ROP = ∠POM.

4. OR = OM are radii of the same circle.

5. In triangles ΔROP and ΔPOM: 

OR = OM   ...(Step 4) 

OP = OP   ...(Common) 

And ∠ROP = ∠POM   ...(Step 3)

6. Hence, ΔROP ≅ ΔPOM by SAS congruence.

7. From congruence, PR = PM.

Therefore, PR = PM, as required.