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प्रश्न
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
उत्तर
Let ∠MPR = x
In ΔMPR
∠MPR = 180º- 90º - x
∠MRP = 90º - x
Similarity, in ΔMPQ
∠MPQ = 90º - ∠MPR
= 90º - x
∠MQP = 180º - 90º - (90º - x)
∠MQP = x
In ΔQMP and ΔPMR
∠MPQ = ∠MRP
∠PMQ = ∠RMP
∠MQP = ∠MPR
∴ΔQMP ~ ΔPMR (By AAA Similarity criterion)
`=>(QM)/(PM) = (MP)/(MR)`
=>PM2 = QM x MR
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