Question

In an equilateral $△BAC$, point $D$ is the midpoint of side $BC$. Prove that $4A{D}^2=3B{C}^2$.

Answer 1

Let AB,BC,ACBE equal

to a

D is the midpoint of

BC.

By Pythagoras theorem

$A{B}^2=A{D}^2+B{D}^2$

$a^2=A{D}^2+(a/2{)}^2$

$a^2-\frac{a^2}{4}=A{D}^2$

$\frac{3a^2}{4}=A{D}^2$

But AB =a

$3A{B}^2=4A{D}^2$

$\therefore$ $4A{D}^2=3A{B}^2$