Question

In an equilateral triangle ABC And D is a point on side BC such that BD=1/3BC Prove that 9AD square= 7AB square

Answer 1

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]