Question

in an equilateral triangle ABC the side BC is trisected at D prove that 9ad square is equal to 7ab square​

Answer 1

REFER TO THE ATTACHMENT FOR FIGURE

Given

• ABC is an equilateral triangle
• BC is trisected at D ,,i.e, BD = BC / 3

To prove

• 9 AD² = 7 AB²

Construction

draw AM ⊥ BC which will bisect BC,,i.e, BM = MC

Proof

Δ ABC is equilateral

so, let us take

AB = BC = AC = a

then,

BD = BC / 3 = a / 3

and

BM = BC / 2 = a / 2

In the figure we can see that,

DM = BM - BD

DM = (a / 2) - ( a / 3)

DM = (3 a - 2 a) / 6

DM = a / 6

NOW,

consider Δ AMB

in which ∠ AMB = 90°

so, by pythagoras theorem

AB² = AM² + BM² -----equation (1)

In Δ AMD by pythagoras

AD² = AM² + DM²

that is

AM² = AD² - DM²  ------equation (2)

substituting equation (2) in equation (1)

AB²= ( AD² - DM² ) + BM²

AB² + DM² - BM² = AD²

(a)² + (a/6)² - (a/2)² = AD²

a² + a²/36 - a² / 4 = AD²

( taking LCM in LHS )

( 36 a² + a² - 9 a² ) / 36 = AD²

28 a² / 36 = AD²

7 a² / 9 = AD²

7 a² = 9 AD²

7 AB² = 9 AD²

PROVED.