In a ∆PQR, PR2 = QR2 and M is a point on side PR such that QM PR. Prove that QM2 = PMXMR.


Given,

∆PQR,


PR2 = QR2 and QMPR


Since,


By Pythagoras theorem,


PR2 – PQ2 = QR2


PR2 = PQ2 + QR2



So, ∆PQR is right angled triangle at Q.


In ∆QMR and ∆PMQ,


M = M [each]


MQR = QPM [each equal to 90°-R]


So, by AAA similarity criteria,


∆QMR ∆PMQ


Now, using property of area of similar triangles, we get,



[ Area of triangles = × base × height]


QM2 = PM × RM


Hence proved.


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