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

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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]

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