# In a ∆PQR, N is a

Given,

In ∆PQR,

N is a point on PR, such that QN PR

And PN.NR = QN2

We have,

PN.NR = QN2

PN.NR = QN.QN And

PNQ = RNQ (common angle)

∆QNP ∆RNQ

Then, ∆QNP and ∆RNQ are equiangular.

PQN + RQN = QRN + QPN

⇒ ∠PQR = QRN + QPN …..(ii)

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

In ∆PQR,

PQR + QPR + QRP = 180°

⇒∠PQR + QPN + QRN = 180°

[ QPR = QPN and QRP = QRN]

⇒∠PQR + PQR = 180°

2PQR = 180°

[using Equation (ii)]

⇒ ∠PQR = 180°/2 = 90°

∴ ∠PQR = 90° [each angle is equal to 90° by SAS similarity criteria]

Hence Proved.

Note:

To remember the process first, show that ∆QNP ∆RNQ, by SAS similarity criterion and then use the property that sum of all angles of a triangle is 180°.

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