In $\Delta ABC$, $BE$ is the angle bisector of $\angle ABC$, $AD$ is the median on side $BC$. $AD$ intersects $BE$ at $O$ perpendicularly. If $AD = BE = 4$, find the lengths of each side of $\Delta ABC$.
What I Tried: At first I was having a hard time trying to make a bit of an accurate picture of the problem, and I made this :-
As solving this, I got no idea. Tried angle-chasing for example, if $\angle ABO = \angle DBO = x$ , then the green angles come to be $(90 – x)$ each, and then you have the brown angle to be $(90 + x)$ . You only get that $\Delta ABO \sim \Delta DBO$ , and that gives me no useful information for now.
I don't think I can use Pythagorean Theorem that much because except $AD = BE = 4$ , I have no other side-lengths to proceed. So right now, I am literally out of ideas.
Can anyone help me do this? Thank You!
Best Answer
In $\triangle ABD$, $BD=AB$. $OA=OD=2$
Let $AB=c$, $AC=b$. $BC=a=2c$.
Also $OE=x$. $OB=4-x$
From $$\dfrac{AE}{CE}=\dfrac{BA}{BC}=\dfrac{1}{2}$$ $$AE = \dfrac{b}{3} , CE= \dfrac{2b}{3}$$
From Apollonius theorem,
$$ b^2 + c^2 = 2(4^2 + c^2)$$
$$ \Rightarrow b^2 - c^2 = 32$$
In right $\triangle BOD$, $$ 2^2 + (4-x)^2 = c^2$$
In right $\triangle AOE$, $$ 2^2 + x^2 = \dfrac{b^2}{9}$$
On solving, $x=1$
So $$ ({a,b,c}) = ({2\sqrt{13},3\sqrt{5},\sqrt{13}}) $$