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

**Solution:**

As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

In ΔPQR; ∠QPR = 90° and PM ⊥ QR

In ΔPQR and ΔMQP

∠QPR = ∠QMP = 90°

∠PRQ = ∠MQP (commom angle)

⇒ ΔPQR ~ ΔMQP (AA Similarity)--------(1)

In ΔPQR and ΔMPR

∠QPR = ∠PMR = 90°

∠PRQ = ∠PRM (commom angle)

⇒ ΔPQR ~ ΔMPR (AA Similarity)--------(2)

From (1) and (2)

ΔMQP ~ ΔMPR

PM / MR = QM / PM (corresponding sides of similar triangles are proportional)

⇒ PM^{2} = QM.MR

**Video Solution:**

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

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 2:

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

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Hence proved that (PM)^{2} = QM.MR