Trigonometric Ceva and a hidden isosceles triangle

Angle sum. $\angle A=50^\circ$, $\angle B=20^\circ+x$, $\angle C=30^\circ+y$. The sum is $180^\circ$, so $x+y=80^\circ$, i.e. $y=80^\circ-x$.

Trigonometric Ceva.
$$\sin\angle BAP\cdot\sin\angle CBP\cdot\sin\angle ACP=\sin\angle PAC\cdot\sin\angle PBA\cdot\sin\angle PCB.$$
Here this reads
$$\sin10^\circ\sin x\sin30^\circ=\sin40^\circ\sin20^\circ\sin(80^\circ-x),\quad\text{i.e.}\quad\frac{\sin x}{\sin(80^\circ-x)}=\frac{\sin40^\circ\sin20^\circ}{\sin10^\circ\sin30^\circ}.$$

Uniqueness. On $(0^\circ,80^\circ)$ the function $x\mapsto\frac{\sin x}{\sin(80^\circ-x)}$ is strictly increasing, so there is at most one solution.

The solution is $x=60^\circ$. Using $2\sin20^\circ\sin40^\circ=\cos20^\circ-\cos60^\circ=\cos20^\circ-\frac12$ and $\sin40^\circ-\sin20^\circ=2\cos30^\circ\sin10^\circ$, one verifies the equality. Hence $x=60^\circ$ is the unique solution.

Conclusion. $x=60^\circ$, $y=20^\circ$, so
$$\angle A=50^\circ,\quad\angle B=80^\circ,\quad\angle C=50^\circ.$$
Since $\angle A=\angle C$, we have $BA=BC$.

Lesson. Trigonometric Ceva converts a cevian configuration into one trigonometric equation; monotonicity then shows that a guessed solution is the only one.