What's the value of the angle $x$ in the figure below?

Question

For reference: (Figure without scale ) $AK = 8, JC = 4, BE = 10$ (Answer:$53^o)$

My progress: exterior angle bisector theorem:

$~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{AE}{CE}=\frac{AB}{BC}$

$\triangle ABK \sim \triangle BCJ~(AA):\\ \frac{8}{4}=\frac{K}{BJ}=\frac{AB}{BC} \implies BK = 2JB, AB = 2BC$

Answer 1

You have found that $C$ is midpoint of $AE$. Construct the median $CD$ with $D$ on $AB$.

Since $CD$ joins the midpoints of two sides of $\triangle ABE$, it is parallel to the third side $BE$ and is half its length.

$\triangle CJD$ turns out to be a $3-4-5$ right triangle with $$x = \angle JDC = \sin^{-1} \frac{4}{5} \approx 53^\circ$$

Answer 2

By trigonometry:

$10=\frac{2bc \sin (\frac{\angle A}{2})}{b-c} \iff 5(b-c) = bc.\sin (\frac{\angle A}{2})\tag1$

$b.\sin (A) = 8\tag2$

$c.\sin (A) = 4\tag3$

(2) and (3) and (1):

$20 \sin^2(\angle A) = \sin (\angle A) \sin(\frac {\angle A}{2}) \cdot 32\\ \implies 5 \sin(\angle A) = 8 \sin(\frac{\angle A}{2})$

$5 \cos (\frac {\angle A}{2}) = 4\implies \frac{\angle A}{2} = 37^\circ \therefore x = 53^\circ$

Answer 3

To show $AB = 2 BC$, you could have equated the area of $\triangle ABC$ as well. Also, here is a construction that helps. Draw perp from $E$ to $BC$ extend.

As $AC = CE$, $\triangle ACK \cong \triangle ECF$ (by A-S-A)

It follows that $EF = AK = 8$. So, $\triangle BFE$ is $3:4:5$ right triangle and hence $x = 53^\circ$.