Geometry - Find angle ∠C of a triangle given angle bisector and two segment ratios. In the triangle ABC, one has $∠A = 70^o$. 
The point $D$ is chosen on the segment
$AC$ such that the angle bisector $AE$ intersects the segment $BD$ at the point $H$, with the following ratios
$$\dfrac{AH}{HE}=\dfrac 31, \>\>\> \dfrac{BH}{HD}=\dfrac 53.$$ 
Find the angle $∠C$.


Examine the two ratios as follows,
$$\frac{5}{3}= \frac{BH}{HD} = \frac{\triangle ABE} {\triangle AED} $$
$$=\frac{\frac{BE}{BC}\triangle ABC}{\frac{EC}{BC}\frac{AD}{AC}\triangle ABC}= \frac{BE}{EC}\left(\frac{CD}{AD}+1\right)\tag{1}$$
Similarly,
$$\frac{3}{1} = \frac{AD}{CD}\left(\frac{EC}{BE}+1\right)\tag{2}$$
Combine (1) and (2) to obtain
$$ \frac{BE}{EC} = 1$$
Since AE is the angle bisector, $\triangle ABE$ and $\triangle AEC$ are congruent, and therefore $\triangle ABC$ is an isosceles triangle, which yields,
$$∠C= \frac{1}{2}(180^\circ -70^\circ) = 55^\circ$$
