Question

triangle
АВС
is an equalateral triangle. Point-D is
on side BC
such that BD = 1/3 Bc. Prove that
25 AD square =21 AB?​

Answer 1

Take another point E in BC , such that AE is perpendicular to BC.

Since it's a perpendicular line in an equilateral ∆ AE divides the∆ into two equal right angle ∆s ABE and ∆ AEC

and hence, BE= EC

BC= BE+BE= 2BE

BE= 1/2 BC

And BE= BD+ED

1/2BC=1/3BC+ED

ED= BC/2 - BC/3= BC/6

Also, we know that AB=BC=CA (sides of equilateral ∆)

let x represent all these sides,

In ∆ ABE

x²= BE²+EA²

EA²= x²-BE²

Now in ∆ ADE

AD²=ED²+AE²

AD²= (x/6)²+(x²-(x/2)²

AD²=x²/36+(x²-x²/4)

AD²= x²/36+[(4x²-x²)/4]

AD²= x²/36+3x²/4

AD²= (x²+27x²)/36

36AD²= 28x²

36AD²= 28AB²

9AD²=7AB²

now multiplying both with 3

27AD²=21AB²