Question

In a ∆PQR, PR² – PQ² = QR² and M is a point on side PR such that QM ⊥ PR. Prove that QM² = PM × MR.

Answer 1

In ∆PQR,

PR² = QR² and QM⊥PR

Using Pythagoras theorem, we have,

PR² = PQ² + QR²

∆PQR is right-angled triangle at Q.

From ∆QMR and ∆PMQ, we have,

∠M = ∠M

∠MQR = ∠QPM   ......[= 90°-∠R]

So, using the AAA similarity criteria,

We have,

∆QMR ∼ ∆PMQ

Also, we know that,

Area of triangles = `1/2` × base × height

So, by property of area of similar triangles,

⇒ `(ar(∆QMR))/(Ar(PMQ)) = (QM)^2/(PM)^2`

⇒ `(ar(∆QMR))/(Ar(PMQ)) = (1/2 xx RM xx QM)/(1/2 xx PM xx QM)`

⇒ `(ar(∆QMR))/(ar(PMQ)) = (QM)^2/(PM)^2`

QM² = PM × RM

Hence proved.