Question

In an equilateral triangle ABC, D is a point on the side BC such that BD = 1 3 BC. Value of 9AD^2 equals​

Answer 1

Answer:

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

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,

AD2 = AE2 + DE2 (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,

AD2 = (√3/2 BC)2 + (BE - BD)2

AD2 = 3/4 BC2 + [BC/2 - BC/3]2

AD2 = 3/4 BC2 + (BC/6)2

AD2 = 3/4 BC2 + BC2/36

AD2 = (27BC2 + BC2) / 36

36AD2 = 28BC2

9AD2 = 7BC2

9AD2 = 7AB2 [Since AB = BC = CA]