Ratio of angles in a triangle came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=AC\neq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 \times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated. 
 A: Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $\Delta APQ$ gives
$$
AQ = AP\cdot\frac{\sin(180^\circ-2x)}{\sin x}
=AP\cdot\frac {\sin(2x)}{\sin x}
=AP\cdot 2\cos x
=2\cos x
\ .
$$
The sides of $\Delta ABC$ are thus $BC=1$ and $1+2\cos x$, and the last again. The sine theorem gives again
$$
\frac {\sin x}1
=\frac{\sin(90^\circ-x/2)}{1+2\cos x}\ .
$$
This gives an equation in $x$ to be solved. 
We rewrite it equivalently step for step:
$$
\begin{aligned}
\sin x &=\frac{\cos(x/2)}{1+2\cos x}\ ,\\
\sin x(1+2\cos x) &= \cos\frac x2\ ,\\
\sin x + \sin 2x &=\cos\frac x2\ ,\\
2\sin \frac{3x}2\cos \frac x2 &= \cos\frac x2\ ,\\
2\sin \frac{3x}2 &= 1\ ,\\
\sin \frac{3x}2 &= \frac 12\ ,\\
\frac{3x}2 &= 30^\circ\ ,\\
x &= 20^\circ\ .
\end{aligned}
$$
Now we have the angles in the triangle(s).
$\widehat{APQ}$ has measure $180^\circ-2x=140^\circ$.
The angles at the base are both $80^\circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
A: Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.

*

*$BCDQ$ is a rhombus;


*$\triangle APQ \cong \triangle PDC$;


*$\triangle PQD$ is an Equilateral triangle;


*$\angle A=20^o$


*$\cdots$

A: I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...


We start with the triangle $\Delta ABC$, and points $P\in AC$, $Q\in AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC\ .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $\Delta ABC$, $\Delta ACB$, $\Delta BAD$, $\Delta BDA$ in different fashions,

$$APQBC \equiv AP'SCB \equiv BQRDA \equiv BQ'P'AD\ ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $\Delta ABC$.
Then the triangles $\Delta APD$ and $\Delta Q'BC$ are equilateral and $x=20^\circ$.

Proof: The triangles $\Delta PQ'B$ and $\Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $A\leftrightarrow B$, and $C\leftrightarrow D$, so by construction we also have the correspondences $P'\leftrightarrow  Q$, $P\leftrightarrow Q'$, $R\leftrightarrow S$. In particular, $PQ\| AB$ (both being $\perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $\angle ABD$, so $$AP=PD\ .$$
Form this, $AP=PD=DA$, so $\Delta APD$ is equilateral. (Same also for $\Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^\circ$, in the sum we consider twice an angle of $60^\circ$ from $\Delta APD$ and further $3x$. From here, we get $x=20^\circ$.
$\square$
