Croatia Mathematical Competitions

Let $D$ be the point on the side $\overline{BC}$ such that $\angle DAC = 90^\circ$. Let us denote $\varphi = \angle CDA$ and $x = |CD|$. Then $\cos \angle ACB = \sin \varphi$.

Then we have $|AC| = x \sin \varphi$ and $|BD| = |AD| = x \cos \varphi$. We also have $\angle BAD = \frac{\varphi}{2}$ and $|AB| = 2x \cos \varphi \cos \frac{\varphi}{2}$.

Since $|BC| + |AC| = 2|AB|$, we get
$$1 + \cos \varphi + \sin \varphi = 4 \cos \varphi \cos \frac{\varphi}{2}.$$
By squaring both sides of the equation we get
$$1 + \cos^2 \varphi + \sin^2 \varphi + 2 \cos \varphi + 2 \sin \varphi + 2 \sin \varphi \cos \varphi = 16 \cos^2 \varphi \cos^2 \frac{\varphi}{2},$$
and further on
$$\begin{aligned} 2(1 + \cos \varphi)(1 + \sin \varphi) &= 8 \cos^2 \varphi(1 + \cos \varphi), \\ 1 + \sin \varphi &= 4(1 - \sin^2 \varphi), \\ (4 \sin \varphi - 3)(\sin \varphi + 1) &= 0, \\ \sin \varphi &= \frac{3}{4}, \end{aligned}$$
where we used $\cos \varphi \neq -1$ and $\sin \varphi \neq -1$, which is valid because the angle $\varphi$ is acute.

Hence, $\cos \angle ACB = \frac{3}{4}$.