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

According to the question,

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.