<s>[INST] 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. 
 [/INST] 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.
</s>