Q) In a Δ PQR, N is a point on PR, such that QN is to PR. If PN x NR = QN², prove that ∠ PQR = 90⁰.

Ans: 

Ans:  Given that, PN x NR = QN²

Therefore,

In Δ PNQ and Δ QNR,

∠ QNP = ∠ RNQ   (given that QN is to PR)

  Δ PNQ Δ QNR

or we can simplify it as, ∠ P = ∠ RQN and  ∠ NQP = ∠ R

By adding these two, we get:

∠ P + ∠ R  = ∠ RQN  + ∠ NQP

Since ∠ RQN  + ∠ NQP = ∠ Q   (part of the same angle)

∠ P + ∠ R  = ∠ Q   ………………………. (i)

Now in Δ PQR, ∠ P + ∠ R + ∠ Q = 180⁰    ………… (ii)

By comparing these equations (i) and (ii), we get:

2 ∠ Q  = 180⁰

or ∠ Q   = ∠ PQR = 90^{0  ………….} Hence proved.