E x a m p l e 9. The angle between the sides of a triangle, the lengths of which are $a$ and $b$, is $\varphi$. Let's find the length of the segment of the bisector of this angle, lying inside the triangle.

G i v e n (Fig. 11): $\triangle A B C,|A C|=b,|B C|=a, \widehat{A C B}=\varphi$, $\widehat{A C D}=\widehat{B C D}$

F i n d: $|C D|$.

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Solution. Let $|CD| = x$. Since $S_{\triangle ABC} = S_{\triangle ACD} + S_{\triangle BCD} = \frac{1}{2} b x \sin \frac{\phi}{2} + \frac{1}{2} a x \sin \frac{\varphi}{2} = \frac{1}{2} x(a+b) \cdot \sin \frac{\varphi}{2}$ and $S_{\triangle ABC} = \frac{1}{2} a b \sin \varphi$, we obtain the following equation:

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-024.jpg?height=317&width=443&top_left_y=230&top_left_x=1252)

Fig. 11

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-024.jpg?height=597&width=462&top_left_y=638&top_left_x=1251)

Fig. 12

Since $\sin \frac{\varphi}{2} \neq 0$, we have $x = \frac{2 a b}{a+b} \cos \frac{\varphi}{2}$. By the problem's condition, $x > 0$. The solution to the equation satisfies this condition.

Example 10. A point on the hypotenuse of a right triangle, equidistant from its legs, divides the hypotenuse into segments of lengths 40 cm and 30 cm. Find the legs of the triangle.

Given (Fig. 12): $\triangle ABC, \widehat{ACB} = 90^\circ, D \in [AB], |AD| = 40 \text{ cm}$, $|BD| = 30 \text{ cm}$, point $D$ is equidistant from $[AC]$ and $[BC]$.

To find: $|AC|$ and $|BC|$.

Solution. Since point $D$ is equidistant from the legs $AC$ and $BC$, $[CD)$ is the angle bisector of $\angle ACB$. Therefore, $\frac{|AC|}{|BC|} = \frac{40}{30} = \frac{4}{3}$. Since $|AB| = 70 \text{ cm}$, let $|AC| = x$, then $|BC| = \sqrt{70^2 - x^2}$, and we can write the equation:

$$
\frac{x}{\sqrt{70^2 - x^2}} = \frac{4}{3}
$$

from which $x_1 = 56, x_2 = -56$.

By the problem's condition, $0 < x < 70$. Only $x_1$ satisfies this condition. Thus, $|AC| = 56 \text{ cm}$, and then $|BC| = 42 \text{ cm}$.

Now let's consider an example where an auxiliary unknown is introduced, and it is eliminated in the process of solving.