PQR is a triangle right angled at P and M is a point on QR such that PM QR. Show that PM2 = QM×MR

Let, MPR = x



In triangle MPR,


MRP = 180o – 90o – x


MRP = 90o – x


Similarly,


In triangle MPQ,


MPQ = 90o - MPR


∠MPQ = 90o – x


In triangle QMP and PMR,


MPQ = MRP


PMQ = RMP


MQP = MPR


Therefore,


ΔQMP ~ΔPMR (By Angle-Angle-Angle similarity)



PM2 = QM×MR

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