Question:
In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$ prove that $\mathrm{AD}^{2}=3 \mathrm{BD}^{2}$.
Solution:
We have to prove that $A D^{2}=3 B D^{2}$.
In right angled $\triangle A B D$, using Pythagoras theorem we get,
$A B^{2}=A D^{2}+B D^{2}$...(1)
We know that in an equilateral triangle every altitude is also median.
Therefore, AD bisects BC.
Therefore, we have $B D=D C$
Since $\triangle A B C$ is an equilateral triangle, $A B=B C=A C$
Therefore, we can write equation (1) as
$B C^{2}=A D^{2}+B D^{2}$....(2)
But $B C=2 B D$
Therefore, equation (2) becomes,
$(2 B D)^{2}=A D^{2}+B D^{2}$
Simplifying the equation we get,
$4 B D^{2}-B D^{2}=A D^{2}$
$3 B D^{2}=A D^{2}$
Therefore, $A D^{2}=3 B D^{2}$.