question about triangle suppose we have some triangle ABC  with AC as base.there is BE bisector and AD median ,they intersect each other  at right angle or are perpendicular,we   should find lengths of triangle.we know that bisector and median are equal  BE=AD=4. from my point of view at suppose level we can conclude that this triangle is equilateral ,because bisector and median are equal,they intersect at right angle or it seems they are perpendicular bisector or altitude?am i correct?also i think that key to solve such problem when there is not given additional information it to suppose such kind of possibilities,please help me to  solve this problem
 A: Hint: The triangle is not equilateral. 

A: Hint: Note that $BA=BD$, so $BC=2BA$.  What familiar type of triangle does this remind you of?
A: I think that what was provided in the discussion under @Andre's hint leads to a solution. (Probably there is a much simpler way, but I'll post my attempt anyway.) 
I will denote by $M$ the intersection of $AD$ and $BC$.
Fact 1 |AB|=|BD|, |AM|=|MD| (and also |AE|=|ED|)
Fact 2  $|\triangle ABE|=|\triangle BDE|=|\triangle CDE|=\frac13|\triangle ABC|$
Fact 3 $|AE|=\frac{|AC|}3$
Fact 4 $|\triangle AME|=\frac{|\triangle AMC|}3=\frac{|\triangle ADC|}6=\frac{|\triangle ABC|}{12}=\frac{|\triangle ABE|}4$
Fact 5 $|EM|=\frac{|EB|}4=1$
As now I know all the lengths of AM, BM,DM, EM, I can use right triangles to calculate:
$|AB|=\sqrt{13}$
$|AE|=\sqrt{5}$ $\Rightarrow$ $|AC|=3\sqrt{5}$
$|BD|=\sqrt{13}$ $\Rightarrow$ $|BC|=2\sqrt{13}$

Now I can use cosine law for triangles ABE and CBE to check, whether I get the same value in both cases. (The values of the cosine should be the same, since BE is the bisector.) I get:
$\cos\frac\beta2 = \frac{13+16-5}{2.4.\sqrt{13}} = \frac{24}{8.\sqrt{13}} = \frac 3{\sqrt{13}}$
$\cos\frac\beta2 =  \frac{4.13+16-4.5}{2.4.2\sqrt{13}} = \frac{48}{16.\sqrt{13}} = \frac 3{\sqrt{13}}$.

You can see why the equalities from Fact 1 hold in Andre's answer and the comments following it.
