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 $∠C$.
On
Since we are only conerning angles, let $Area(ΔAHB) = 15$
Area of triangle = ${1\over2}\,a\,b\,\sin(C)$.
If side b changed to $(b + k\;b) \text{, added area} = k\;\{ {1\over2}\,a\,b\,\sin(C)\}$
$\begin{matrix} {HD \over BH}={3 \over 5} &→ Area(ΔAHD) = {3 \over 5}(15) = 9 \cr {HE \over AH}={1 \over 3} &→Area(ΔEHB) = {1 \over 3}(15) = 5 \cr &→ Area(ΔEHD) = {1 \over 3}(9) = 3 \cr \end{matrix}$
Let $k = \large{CE \over BE}$
$\begin{align} Area(ΔCDE) &= k\;Area(ΔBDE) \cr &= (k)(5+3)\cr\cr Area(ΔCDE) &= k\;Area(ΔAEB) - Area(ΔAED) \cr &= (k)(15+5) - (9+3) \cr\cr 8k &= 20k-12 \cr k &=1 \end{align}$
This implied $CE=BE,\quad ΔAEB ≅ ΔAEC$
$$∠C = {180° - 70° \over 2} = 55°$$
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,
$$∠C= \frac{1}{2}(180^\circ -70^\circ) = 55^\circ$$