ABC is a triangle and the line YCX is parallel to AB such that AX and BY are the angular bisectors of angle A and I'm interested in the following problem:

$ABC$ is a triangle and the line $YCX$ is parallel to $AB$ such that $AX$ and $BY$ are the angular bisectors of angle $A$ and angle $B$ respectively. If $AX$ meets $BC$ at $D$ and $BY$ meets $AC$ at $E$ and if $YE =XD$, then prove that $AC=BC$.

My attempt. I tried using some angle chasing then I used cosine formula for finding the two equal segment but that goes too lengthy.
Any help will be highly appreciated!
 A: Since $XY||AB$, in the standard notation we obtain:
$\measuredangle CXD=\frac{\alpha}{2},$ $\measuredangle CYE=\frac{\beta}{2},$ $\measuredangle DCX=\beta$, $\measuredangle ECY=\alpha$, $CX=b$ and $CY=a.$
Thus, by the law of sines for $\Delta DCX$ we obtain:
$$\frac{XD}{\sin\beta}=\frac{b}{\sin\left(\beta+\frac{\alpha}{2}\right)},$$ which gives
$$XD=\frac{b\sin\beta}{\sin\left(\beta+\frac{\alpha}{2}\right)}=\frac{bl_a}{c}.$$
By the same way we can get:
$$YE=\frac{al_b}{c},$$ which gives $$al_b=bl_a.$$
Now, $$l_a=\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2}}{2}}}{b+c}=\frac{2\sqrt{(a+b+c)(b+c-a)bc}}{b+c}.$$
By the same way:
$$l_b=\frac{2\sqrt{(a+b+c)(a+c-b)ac}}{a+c}.$$
Thus, $$a^3(a+c-b)(b+c)^2=b^3(b+c-a)(a+c)^2.$$
Now, let $a>b$.
Thus, $$a+c-b>b+c-a,$$
$$a^2(b+c)^2>b^2(a+c)^2,$$ which gives $$a^3(a+c-b)(b+c)^2>b^3(b+c-a)(a+c)^2,$$
which is a contradiction.
By the same way if $a<b$ so
$$a^3(a+c-b)(b+c)^2<b^3(b+c-a)(a+c)^2,$$ which is a contradiction again.
Id est, $a=b$ and we are done!
