# ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC (ii) MD ⊥ AC

(iii) CM = MA = 1/2 AB

**Solution:**

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle that is parallel to another side bisects the third side.

Let's construct the triangle according to the question.

**(i)** In ΔABC,

It is given that M is the mid-point of AB and MD || BC.

∴ D is the mid-point of AC. [Converse of mid-point theorem]

**(ii**) As DM || CB and AC is a transversal,

∠MDC + ∠DCB = 180° [Co-interior angles]

∠MDC + 90° = 180°

∠MDC = 90°

∴ MD ⊥ AC

**(iii)** Join MC

In ΔAMD and ΔCMD,

AD = CD (D is the mid-point of side AC)

∠ADM = ∠CDM (Each 90°)

DM = DM (Common)

∴ ΔAMD ≅ ΔCMD (By SAS congruence rule)

Therefore, AM = CM (By CPCT)

However, we also know that AM = 1/2 AB (M is the mid-point of AB)

∴ CM = AM = 1/2 AB

**Video Solution:**

## ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

i) D is the mid-point of AC ii) MD ⊥ AC iii) CM = MA = 1/2 AB

### NCERT Maths Solutions Class 9 - Chapter 8 Exercise 8.2 Question 7:

**Summary:**

If ABC is a triangle right angled at C and a line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D, then D is the mid-point of AC, MD ⊥ AC, and CM = MA = 1/2 AB.