# 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:**

Let's construct a diagram according to the given question.

We know that 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 each other and to the whole triangle.

In ΔPQR we have,

∠QPR = 90° and PM ⊥ QR

In ΔPQR and ΔMQP

∠QPR = ∠QMP = 90°

∠PQR = ∠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 equation (1) and (2)

ΔMQP ~ ΔMPR

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

⇒ PM^{2} = QM.MR

Hence proved.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 2

**Summary:**

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.

**☛ Related Questions:**

- In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB² = BC.BD (ii) AC² = BC.DC (iii) AD² = BD.CD.
- ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC^2.
- ABC is an isosceles triangle with AC = BC. If AB^2 = 2AC^2, prove that ABC is a right triangle.
- ABC is an equilateral triangle of side 2a. Find each of its altitudes.

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