Question

PQR is triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM ^2 = QM×MR.​

Answer 1

Given:-

• PQR is a right angled at a P and M is a point on QR such that PM ⊥ QR.

To prove:-

• PM = QM × MR

Solution:-

→ ∠P = 90°

→ ∠1 + ∠2 = 90° --(i)

→ ∠M = 90°

According to the question:-

In ∆PMQ

→ ∠1 + ∠3 + ∠5 = 180°

→ ∠1 + ∠3 + 90° = 180°

→ ∠1 + ∠3 = 180° - 90°

→ ∠1 + ∠3 = 90° ---(ii)

Sum of (i) & (ii)

→ ∠1 + ∠2 = ∠1 + ∠3

→ ∠2 = ∠3

In ∆ QPM & ∆RPM

→ ∠3 = ∠2

→ ∠5 = ∠6 ( Each of 90° )

By AD similarity,

→ ∆QPM ~ ∆RPM

→ ar(∆QMP)/ar(∆RPM) = PM^2/MR^2

→ 1/2 × QM × PM / 1/2 × RM × PM = PM^2/MR^2

→ QM/RM = PM^2/MR^2

→ PM = QM × RM.

Hence proved.