# In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC² = AD² + BC.DM + (BC/2)²

(ii) AB² = AD² - BC.DM + (BC/2)²

(iii) AC² + AB² = 2AD² + 1/2BC²

**Solution:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Theorem 6.8

(i) In ΔAMC

∠AMC = 90°

AC² = AM² + CM²

= AM² + [DM + CD]²

= AM² + DM² + CD² + 2DM.CD

= AD² + (BC/2)² + 2DM (BC/2)

Since, in ΔAMD, AD² = AM² + DM² and D is the midpoint of BC means BD = CD = BC/2

AC² = AD² + BC.DM + (BC/2)² ......(i)

(ii) In ΔAMB

∠AMB = 90°

AB² = AM² + BM²

= AM² + [BD - DM]²

= AM² + BD² + DM² - 2BD.DM

= AM² + DM² + (BC/2)² - 2 (BC/2)DM

Since, in

ΔAMD, AD² = AM² + DM² and D is the midpoint of BC means BD = CD = BC/2

AB² = AD² - BC.DM + (BC/2)^{2} .....(ii)

(iii) Adding (i) and (ii)

AC² + AB² = AD² + (BC/2) + BC.DM + AD² + (BC/2) - BC.DM

AC² + AB² = 2AD² + 2(BC/2)²

AC² + AB² = 2AD² + 1/2 BC²

**Video Solution:**

## In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that : (i) AC² = AD² + BC.DM + (BC/2)² (ii) AB² = AD² - BC.DM + (BC/2)² (iii) AC² + AB² = 2AD² + 1/2BC²

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 5:

In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that : (i) AC² = AD² + BC.DM + (BC/2)² (ii) AB² = AD² - BC.DM + (BC/2)² (iii) AC² + AB² = 2AD² + 1/2BC²

In the above figure, AD is a median of a triangle ABC and AM ⊥ BC. Hence proved that AC² = AD² + BC.DM + (BC/2)² and AB² = AD² - BC.DM + (BC/2)² and AC² + AB² = 2AD² + 1/2BC²