**Q) **In a Δ PQR, N is a point on PR, such that QN is to PR. If PN x NR = QN^{2}, prove that ∠ PQR = 90^{0}.

**Ans: **

**Ans: **Given that, PN x NR = QN^{2}

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 = 180^{0} ………… (ii)

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

2 ∠ Q = 180^{0}

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