Given: In △PQR, ∠P = 90° and PM ⊥ QR.
R.T.P : PM2 = QM . MR
Proof: In △PQR; △MPR
∠P = ∠M [each 90°]
∠R = ∠R [common]
∴ △PQR ~ △MPR ……… (1)
[A.A. similarity]
In △PQR and △MQP,
∠P = ∠M (each 90°)
∠Q = ∠Q (common)
∴ △PQR ~ △MQP ……… (2)
[A.A. similarity]
From (1) and (2),
△PQR ~ △MPR ~ △MQP [transitive property]
∴ △MPR ~ △MQP
\(\frac{MP}{MQ}=\frac{PR}{QP}=\frac{MR}{MP}\)
[Ratio of corresponding sides of similar triangles are equal]
\(\frac{PM}{QM}=\frac{MR}{PM}\)
PM . PM = MR . QM
PM2 = QM . MR [Q.E.D]