finding the base of a triangle In A triangle ABC ,AB=AC.D is a point inside the triangle such that AD=DC.Median on AC from D meets median on BC  from A at the centroid of the triangle.If the area of the triangle  ABC equals to $4\sqrt 3$ .Find the base i.e. BC.
The method that i have used to solve this problem works by ending with  two answers for $\frac12$ of BC and then having to check which one works by trial and error.It is more or less efficient,but I am assuming there's a better one.
 A: If $AD=DC$ ike you said then $\Delta ABC$ is isosceles. Also we have that triangle $\Delta ABC$ is isosceles. A centroid is the intersection of the three medians of a triangle intersecting opposite vertices thus if the median from $AC$ meets median on $BC$ at the centroid, then both medians must intersect their respective opposite vertices. Then we can conclude that the median from $D$ on $AC$ is also a median from $B$ on $AC$.  This median forms an angle of $90°$ with side $AC$ since $\Delta ADC$ is isosceles. Now we know that triangle $\Delta ABC$ has two medians that are also perpendicular bisectors, it is easy to see that triangle $\Delta ABC$ is equilateral and $AB = BC = CA. $
$$ [\Delta ABC]= \cfrac 12 BC^2\sin 60° = \cfrac{\sqrt 3}{4}BC^2=4\sqrt 3 \implies BC=4$$
A: This is my own solution.At first,we know that  triangle ABC is isosceles.We also know that  tri ADC is also isosceles.So the median of that triangle bisects AC in a right angle. Note that the side opposite angle B is AC,the same as angle D.The median from angle B , like that of  angle D will also meet at the centroid.So we can draw the inference that  median from angles B and D are actually the same straight lline.But then the median from angle B bisects AC perpendicularly.So AB =BC=AC (since median of an isosceles triangle bisects it  perpendicularly.This proves that ABC is an equilateral triangle.Now let the median on BC bisect BC at M.Then ,by congruency criterion,triangles ABM and AMC are equal, and so area of  each of them is 2× sqrt 3.
We know,
$2\times \sqrt{3} = 1/2(2\times 2\times \sqrt{3})$. So MC equals either (\sqrt{2} )\times 3 or simply 2.Now let us use the pythagorean theorem to see which fits.It seems that 2 is the only solution for MC.so $BC =4$.That is how I solved it.
