Finding the $x$ and $y$ part of the side of the triangle

Question

I need help finding x and y in this triangle. Conditions: It is not a right triangle; there are no given angles; $u$ doesn't bisect the corresponding angle; $u$ doesn't split $c$ in two equal parts; $c,b,a$ and $u$ are given; Triangle example

Answer 1

Check someone did suggest Stewart's Theorem as a comment!

$xb^2 + yc^2 = (x+y)(u^2+xy)=au^2+axy$

From here, $x$ can be found out this way: \begin{align} & x(b^2-ay)+yc^2=au^2 \\ & \implies x(b^2-ay-c^2)=a(u^2-c^2) \\ & \implies \boxed{x = \frac{a(u^2-c^2)}{b^2-ay-c^2}} \end{align}

Similarly, $$\boxed{y=a-x =a- \frac{a(u^2-c^2)}{b^2-ay-c^2}}$$