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

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    $\begingroup$ Lookup Stewart's theorem. $\endgroup$
    – dxiv
    Feb 17, 2018 at 23:24
  • 1
    $\begingroup$ Works perfect. Thank you $\endgroup$
    – adl
    Feb 18, 2018 at 0:12

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

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

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