Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?
$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $\angle C$ is divided by $\overline{CP}$. Also its clear that $\overline{BP}$ is perpendicular to leg $\overline{AC}$.
On
Consider the figure below.
$\\$Thanks to @saulspatz answer, and based on Trigonometric Form of Ceva's Theorem we may write the following:
$$Sin(\angle B1).Sin(\angle C1).Sin(\angle A1) = Sin(\angle B2).Sin(\angle C2).Sin(\angle A2)$$
$$\\$$so we have:
$$\frac{Sin(\angle C1)}{Sin(\angle C2)} = \frac{Sin(40^o)*Sin(20^o)}{Sin(10^o)*Sin(60^o)}$$
$\\$Also we have $\angle C1 + \angle C2 = 50^o$, and by solving for $\angle C2$:
$$\angle C2 = 20^o$$
$\\$Hence: $$\angle x = 100^o$$
Construct equilateral triangle $ABX$ as in the picture.
Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.
On the other hand, $AB=AX=AC$, so $A$ is the circumcenter of triangle $BXC$, hence $\angle CXB = \frac 12 \cdot \angle CAB = \frac 12 \cdot 80^\circ = 40^\circ$.
Since $\angle PXB = 40^\circ = \angle CXB$, points $X, P, C$ are collinear.
Since $AC=AX$, we have $\angle PCA = \angle AXP = 2\cdot \angle ABP = 20^\circ$, hence $$\angle APC = 180^\circ - \angle CAP - \angle PCA = 180^\circ - 60^\circ - 20^\circ = 100^\circ.$$