Question

In an equilateral $\Delta$ ABC, D is a point on side BC such that BD$=\frac{1}{3}$BC. Prove that $9(AD{)}^2=7(AB{)}^2$.

Answer 1

Let $E$ be the midpoint of BC so that $BE=EC=\frac{BC}{2}$

and the length of the sides of the equilateral triangle is $x$. So, we have

$AB=BC=CA=x\dots \left(1\right)$

$BE=EC=\frac{x}{2}$

And it is given that $BD=\frac{BC}{3}=\frac{x}{3}$

also $BE=\frac{x}{2}$

$\therefore DE=BE-BD=\frac{x}{2}-\frac{x}{3}=\frac{x}{6}$

Now, In the right angle $△AED$

we have $AE=\frac{\sqrt{3}x}{2}$

$A{D}^2=A{E}^2+D{E}^2=\frac{3x^2}{4}+\frac{x^2}{36}=\frac{28x^2}{36}$

$36A{D}^2=28x^2$

Now put $x^2=A{B}^2$ from equation (1), we get

$36A{D}^2=28A{B}^2$

$⟹9A{D}^2=7A{B}^2$ $\left[henceproved\right]$