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