2. Let $D$ be the midpoint of side $AB$, $E$ the intersection of side $BC$ and the angle bisector of $\angle BAC$, and $F$ the orthogonal projection of point $E$ onto side $AB$ of triangle $ABC$. Suppose that $\angle CDA = \angle ACB$ and $|CE| = |BF|$. Determine the measures of the angles of triangle $ABC$.
II/2.

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-20.jpg?height=334&width=531&top_left_y=2149&top_left_x=271)

Since $\angle C D A = \angle A C B$, triangles $A D C$ and $A C B$ are similar, so

$$
\frac{|A C|}{|A B|} = \frac{|A D|}{|A C|} = \frac{|A B|}{2|A C|}
$$

from which it follows that $|A B| = |A C| \sqrt{2}$. Since $A E$ is the angle bisector of $\angle B A C$, we have

$$
\frac{|B E|}{|C E|} = \frac{|A B|}{|A C|} = \sqrt{2}
$$

and from $|C E| = |B F|$ it follows that $|B E| = \sqrt{2} |B F|$. Since $E F B$ is a right triangle with a right angle at vertex $F$, by the Pythagorean theorem, it follows that $|E F| = |B E| \sqrt{2}$, which means that triangle $E F B$ is also isosceles and $\angle C B A = \angle E B F = \frac{\pi}{4}$. We already know that triangles $A B C$ and $A C D$ are similar, so $\angle A C D = \angle C B A = \frac{\pi}{4}$. Let $D^{\prime}$ be the orthogonal projection of point $A$ onto line $C D$. Then $A D^{\prime} C$ is an isosceles right triangle, so $\left|A D^{\prime}\right| = \frac{|A C|}{\sqrt{2}}$. On the other hand, we have already calculated that $|A D| = \frac{1}{2}|A B| = \frac{|A C|}{\sqrt{2}}$. Since points $D$ and $D^{\prime}$ both lie on line $C D$, and $D^{\prime}$ is the orthogonal projection of point $A$ onto $C D$, it follows that $D^{\prime} = D$. Therefore, $\angle C D A = \frac{\pi}{2}$. It follows that $\angle A C B = \frac{\pi}{2}$ and $\angle B A C = \frac{\pi}{4}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=60&width=1633&top_left_y=872&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=51&width=1633&top_left_y=931&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=48&width=1633&top_left_y=981&top_left_x=220)

Projecting point $A$. ....................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_8c492ac180897ee339acg-21.jpg?height=54&width=1631&top_left_y=1075&top_left_x=224)

2nd method. As in the first solution, we show that $|A B| = \sqrt{2} |A C|$ and $\angle C B A = \frac{\pi}{4}$. From this, by the cosine rule, it follows that

$$
|A C|^{2} = |A B|^{2} + |B C|^{2} - 2 |A B| |B C| \frac{\sqrt{2}}{2} = 2 |A C|^{2} + |B C|^{2} - 2 |A C| |B C|
$$

which can be rewritten as $(|A C| - |B C|)^{2} = 0$. Therefore, $|A C| = |B C|$ and thus $\angle B A C = \angle C B A = \frac{\pi}{4}$ and $\angle A C B = \frac{\pi}{2}$.