Let $\Delta ABC$ be an isosceles triangle, s.t. $\overline{AC}$ and $\overline{BC}$ have the same length. Let $D$ be some point on $\overline{BC}$, s.t. $\overline{BD}$ and $\overline{CD}$ have the same length. Assume that $\Delta ABD$ is also an isosceles triangle, s.t. $\overline{AB}$ and $\overline{AD}$ have the same length.
Find the size of the angle $\angle ACB$. (Marked red in the sketch)
My idea to make use of the assumption $\left| \overline{BD} \right| = \left| \overline{CD} \right|$ was to add a copy of the triangle $\Delta ABD$ to the outside of $\overline{CD}$. One receives a parallelogram $ABA'C$, but I cannot seem to make use of it. If I recall correctly, an elementary solution exists and trigonometry is not necessary, but I have failed to find it for much time now...
Answer: Exact form is $\angle ACB = \arccos \dfrac34$ and approximate form is $\angle ACB \approx 41,4^\circ$.
Let's solve it. Without loss of generality we can assume that $|CD|=|BD|=1$, $|AC|=2$ and $|AD|=|AB|=x$. Since $\triangle ACB \sim \triangle BAD$, $\dfrac2x= \dfrac x2$ and so $x=\sqrt 2$. By cosine theorem in $\triangle ABC$, $$\cos \angle ACB =\dfrac{2^2+2^2-2}{2\cdot 2\cdot 2}=\dfrac 34$$.