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


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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