I need to calculate $PQ$ knowing that $AC = 20$.
This is what I got so far: If I call the point between $P$ and $A$, "$M$" and If I call the angle:
$$\measuredangle{QPB} = y$$
Then:
$$\measuredangle{BMA} = \measuredangle{PMC} = y$$
And if we call the point between $C$ and $P$, "$N$", then:
$$ \measuredangle{NCP} = \measuredangle{NPC} = x$$
So there are two isosceles triangles $\triangle CNP$ and $\triangle PNM$. $CN = NM$.
I think the triangles $\triangle CPB$ and $\triangle CAB$ are congruent because they are both rectangle triangles and possess the same hypotenuse, but I don't know how to prove it. I tried working with the angles and their Cosines, but the it gets way too hard.
Thanks for the help.
Say that $\widehat{ABC}=2x$. Then $$PQ = BP\sin x = BC \cos x\sin x = AC\frac{\sin x\cos x}{\sin(2x)} = \frac {AC}{2}=10.$$
As an alternative, notice that $B,A,P,C$ all lie on the circle $\Gamma$ having $BC$ as a diameter. Since $BP$ bisects $\widehat{ABC}$, $P$ is the midpoint of the minor arc $AC$ on $\Gamma$. This gives that $P$, the midpoint $M$ of $AC$ and the midpoint $O$ of $BC$ (that is the center of $\Gamma$) are collinear. If we take $R$ as the symmetric of $P$ with respect to $\Gamma$, we have that $BPCR$ is a rectangle, hence $BPC$ is congruent to $PCR$ and $PQ$ and $CM$ have the same length. This happen also because both $CM$ and $PQ$ are the heights of an isosceles triangle, $OPC$.