## Task A-1.3.

Given is an isosceles triangle $A B C$ with $\overline{B C}$ as the base. On the outer side of this triangle, isosceles triangles $C B D, A C E$, and $B A F$ similar to triangle $A B C$ are drawn, with bases $\overline{B C}, \overline{C A}$, and $\overline{B F}$, respectively. If $\varangle C A B=38^{\circ}$, determine $\varangle E D F$.
## Solution.

In an isosceles triangle, the angles at the base are equal. Let the measure of angle $\varangle C A B$ be $\alpha$, and the measures of angles $\varangle A B C$ and $\varangle B C A$ be $\beta$. Then $2 \beta + \alpha = 180^{\circ}$, from which $\beta = 71^{\circ}$.

We introduce the following notations for the lengths of segments: $b = |A B| = |C A|, a = |B C|, c = |C E|$.

Triangles $A B C$ and $C B D$ are similar isosceles triangles that share the base, so they are actually congruent. Similarly, triangles $A B C$ and $B A F$ are similar isosceles triangles that share one leg, so they are also congruent.

![](https://cdn.mathpix.com/cropped/2024_05_30_a439f3fdcb63d942eea2g-04.jpg?height=811&width=985&top_left_y=888&top_left_x=547)

Therefore, we conclude

$$
|B D| = |D C| = |A B| = b \quad \text{and} \quad |B F| = |B C| = a
$$

From the similarity of triangles $A B C$ and $E A C$, we get

$$
\frac{|C E|}{|A B|} = \frac{|A C|}{|B C|} \Longrightarrow \frac{c}{b} = \frac{b}{a} \Longrightarrow c = \frac{b^{2}}{a}
$$

Consider triangles $C D E$ and $B F D$. Angle $\varangle E C D$ together with angles $\varangle A C E, \varangle B C A$, and $\varangle D C B$ form a full angle. These angles are base angles in isosceles triangles, so we have

$$
\varangle E C D = 360^{\circ} - \varangle A C E - \varangle B C A - \varangle D C B = 360^{\circ} - 3 \beta
$$

Similarly, angle $\varangle D B F$ together with angles $\varangle F B A, \varangle A B C$, and $\varangle C B D$ (which are also base angles in isosceles triangles) form a full angle, so we have

$$
\varangle D B F = 360^{\circ} - 3 \beta = \varangle E C D .
$$

Furthermore, for the ratios of the sides of triangles $C D E$ and $B F D$, we have

$$
\frac{|C E|}{|C D|} = \frac{c}{b} = \frac{\frac{b^{2}}{a}}{b} = \frac{b}{a} = \frac{|B D|}{|B F|}
$$

Thus, by the S-A-S similarity theorem, we conclude that triangles $C D E$ and $B F D$ are similar, and $\varangle C D E = \varangle B F D$. In particular, we have

$$
\varangle F D B + \varangle C D E = \varangle F D B + \varangle B F D = 180^{\circ} - \varangle D B F = 180^{\circ} - (360^{\circ} - 3 \beta) = 3 \beta - 180^{\circ} .
$$

Finally, the measure of the desired angle is

$$
\varangle E D F = \varangle F D B + \varangle B D C + \varangle C D E = \alpha + 3 \beta - 180^{\circ} = \beta = 71^{\circ}
$$