Problem: Let $ABC$ be a triangle, and $M$ an interior point such that
$\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$.
Prove that the triangle is isosceles.

Solution: Let $\theta = \angle MBC < 80\dg$.
By trig Ceva, we get
\[
  \frac{\sin 10\dg}{\sin 40\dg} \cdot \frac{\sin \theta}{\sin 20\dg}
  \cdot \frac{\sin 30\dg}{\sin (80\dg-\theta)} = 1.
\]
This simplifies to
\[ \sin \theta = 4 \sin(80\dg- \theta) \sin 40\dg \cos 10\dg.\]
\begin{claim*}
  We have $\theta=60\dg$.
\end{claim*}
\begin{proof}
  The left-hand side is increasing in $\theta$ and the right-hand
  side is decreasing in $\theta$,
  so at most one value of $\theta$ works.
  But we also have
  \begin{align*}
    4\sin20\dg \sin40\dg \cos10\dg
    &= 2\left( \cos20\dg-\cos60\dg \right) \cos 10\dg \\
    &= 2\cos20\dg \cos 10\dg + \sin 80\dg \\
    &= (\cos 30\dg - \cos 10\dg) + \sin 80\dg = \cos 30\dg
  \end{align*}
  as desired.
\end{proof}