# In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^{2} = 7AB^{2}

**Solution:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [Pythagoras theorem]

In ΔABC;

AB = BC = CA and BD = 1/3 BC

Draw AE ⊥ BC

BE = CE = 1/2 BC

[In an equilateral triangle perpendicular drawn from vertex to opposite side bisects the side]

Now, In ΔADE

AD^{2} = AE^{2} +DE^{2} (Pythagoras theorem)

= (√3/2 BC)^{2} + (BE - BD)^{2}

[AE is the height of an equilateral triangle which is equal to √3/2 side]

AD^{2} = 3/4 BC^{2} + [BC/2 - BC/3]^{2}

AD^{2} = 3/4 BC^{2} + (BC/6)^{2}

AD^{2} = 3/4 BC^{2} + BC^{2}/36

AD^{2} = (27BC^{2} + BC^{2}) / 36

36AD^{2} = 28 BC^{2}

9AD^{2} = 7BC^{2}

9AD^{2} = 7AB^{2} [Since AB = BC = CA]

**Video Solution:**

## In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^{2} = 7AB^{2}

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 15:

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^{2} = 7AB^{2}

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Hence proved that 9AD^{2} = 7AB^{2}