Question

I draw a chord PQ of the circle with the centre A, which is not a diameter. I draw a perpendicular AM on PQ from A. I prove with reason that PM = MQ. [Let me prove it by drawing]​

Answer 1

Answer:

Given :

• AM is the perpendicular on the circle.

To Prove :

• PM = MQ

Construction :

• A, P and A, Q is joined.

Proof :

$\because$ AM $\perp$ PQ

∴ ∆APM and ∆AMQ are right angled triangle.

In right angled triangle ∆APM and ∆AMQ

Hypotenuse AP = Hypotenuse AQ ( radius of the same circle )

AM is the common side

∴ ∆APM ≌ ∆AMQ ( R - H - S condition )

∴ PM = MQ ( Similar side of congruent triangles )

( PROVED )

Answer 2

EXPLANATION.

Draw a chord PQ of the circle with the center A.

Draw perpendicular AM on PQ from A.

As we know that,

PA = QA [ same hypotenuse].

AM ⊥ PQ. [Given].

In ΔAMP = AMQ.

M = Right angle = 90°.

We can write,

⇒ PQ = PM + MQ.

⇒ In ΔAMP,

⇒ AP² = AM² + PM².

⇒ In ΔAMQ,

⇒ AQ² = AM² + QM².

By the [R.H.S congruence].

⇒ ΔAMP ≅ ΔAMQ [ R.H.S].

⇒ PM = MQ. [ Same sides on triangle ].