## Task A-3.1.

In triangle $A B C$, it holds that $|B C|+|A C|=2|A B|$ and $\angle B A C - \angle C B A = 90^{\circ}$.

Determine the cosine of the angle $\angle A C B$.
## First Solution.

Let point $D$ be on side $\overline{B C}$ such that $\varangle C A D=90^{\circ}$. Let $\varphi=\varangle C D A$ and $x=|C D|$. Then $\cos \varangle A C B=\sin \varphi$.

![](https://cdn.mathpix.com/cropped/2024_05_30_6f525b6526cae5162cdfg-10.jpg?height=416&width=788&top_left_y=957&top_left_x=634)

Then $|A C|=x \sin \varphi$ and $|A D|=|D B|=x \cos \varphi$. Also, $\varangle D A B=\frac{1}{2} \varphi$ and $|A B|=$ $2 x \cos \varphi \cos \frac{1}{2} \varphi$.

Due to the condition $|B C|+|C A|=2|A B|$, we have

$$
1+\cos \varphi+\sin \varphi=4 \cos \varphi \cos \frac{1}{2} \varphi
$$

By squaring both sides of the equation, we get

$$
1+\cos ^{2} \varphi+\sin ^{2} \varphi+2 \cos \varphi+2 \sin \varphi+2 \sin \varphi \cos \varphi=16 \cos ^{2} \varphi \cos ^{2} \frac{1}{2} \varphi
$$

and further simplification yields

$$
\begin{aligned}
2(1+\cos \varphi)(1+\sin \varphi) & =8 \cos ^{2} \varphi(1+\cos \varphi) \\
1+\sin \varphi & =4\left(1-\sin ^{2} \varphi\right) \\
(4 \sin \varphi-3)(\sin \varphi+1) & =0 \\
\sin \varphi & =\frac{3}{4}
\end{aligned}
$$

where we used $\cos \varphi \neq-1$ and $\sin \varphi \neq-1$, which holds because $\varphi$ is an acute angle.

Thus, $\cos \varangle A C B=\frac{3}{4}$.