Question

In a triangle PQR, N is a point on PR such that QN ⊥ PR. If PN . NR = QN², prove that ∠PQR = 90°.

Answer 1

Given, ∆PQR,

N is a point on PR, such that QN ⊥ PR

And PN . NR = QN²

To prove: ∠PQR = 90°

Proof: We have, PN . NR = QN²

⇒ PN . NR = QN . QN

⇒ `("PN")/("QN") = ("QN")/("NR")`   ...(i)

In ∆QNP and ∆RNQ,

`("PN")/("QN") = ("QN")/("NR")`

And ∠PNQ = ∠RNQ   ...[Each equal to 90°]

∴ ∆QNP ~ ∆RNQ      ...[By SAS similarity criterion]

Then, ∆QNP and ∆RNQ are equiangulars.

i.e., ∠PQN = ∠QRN

⇒ ∠RQN = ∠QPN

On adding both sides, we get

∠PQN + ∠RQN = ∠QRN + ∠QPN

⇒ ∠PQR = ∠QRN + ∠QPN     ...(ii)

We know that, sum of angles of a triangle is 180°

In ∆PQR,

∠PQR + ∠QPR + ∠QRP = 180°

⇒ ∠PQR + ∠QPN + ∠QRN = 180°   ...[∵ ∠QPR = ∠QPN and ∠QRP = ∠QRN]

⇒ ∠PQR + ∠PQR = 180°    ...[Using equation (ii)]

⇒ 2∠PQR = 180°

⇒ ∠PQR = `180^circ/2` = 90°

∴ ∠PQR = 90°

Hence proved.