## Task B-2.7.

A line passing through the vertex $C$ of the right angle of triangle $A B C$ with the shorter leg forms an angle of $30^{\circ}$ and divides the hypotenuse in the ratio 1 : 2. If the length of the shorter leg is $\sqrt{3}$, determine the length of the hypotenuse.
## First Solution.

Assume that $|A C|$ is the shorter leg of a right triangle. From vertex $A$, draw a line that intersects line $C D$ at point $E$ such that angle $\varangle C A E$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-12.jpg?height=471&width=425&top_left_y=661&top_left_x=747)

Triangle $C A E$ is a right triangle, so we have:

$\cos 30^{\circ}=\frac{|C A|}{|C E|}$, or $\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{|C E|}$, thus $|C E|=2$,

and by the Pythagorean theorem, $|A E|=1$.

(This can also be calculated by considering triangle $C A E$ as half of an equilateral triangle)

Triangles $A D E$ and $C D B$ are similar because they have two equal angles: $\varangle C D B=\varangle A D E$ because they are vertical angles, $\varangle D A E=\varangle C B D$ because they are angles with parallel sides.

Then we have:

$$
|A E|:|B C|=|A D|:|D B|=1: 2 \Rightarrow|B C|=2
$$

Applying the Pythagorean theorem to triangle $A B C$, we get that the length of the hypotenuse is

$$
|A B|=\sqrt{2^{2}+(\sqrt{3})^{2}}=\sqrt{7}
$$