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

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

• Lookup Stewart's theorem.
– dxiv
Feb 17, 2018 at 23:24
• Works perfect. Thank you
Feb 18, 2018 at 0:12

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

• P.S : I know that the doubt's been resolved, yet to increase my points (I'm a new user), I just answered it.
– user532368
Feb 18, 2018 at 3:31
• Your expression for $x$ and $y$ should not have $x$ or $y$ because $x$ and $y$ are not given. Feb 25, 2018 at 18:27
• Yes, $y+x=a$, just replace this and the terms can be removed.
– user532368
Feb 25, 2018 at 18:28
• After we replace $y$ with $a-x$ in the expression of $x$, then we still need to solve for $x$. Feb 25, 2018 at 19:12
• @Delong that can be done by the OP.
– user532368
Feb 25, 2018 at 19:17