# 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:**

We know that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In ΔABC as shown in the figure above,

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

Draw AE ⊥ BC

We know that in an equilateral triangle perpendicular drawn from vertex to opposite side bisects the side

Thus, BE = CE = 1/2 BC

Now, in ΔADE,

AD^{2} = AE^{2} + DE^{2} (Pythagoras theorem) ---------------- (1)

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

Thus, AE = √3/2 BC

Also, DE = BE - BD [From the diagram]

Substituting these in equation(1) we get,

AD^{2 }= (√3/2 BC)^{2} + (BE - BD)^{2}

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} = 28BC^{2}

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

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

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

**Video Solution:**

## In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 15

**Summary:**

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}.

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