[The ratio in which the bisector divides the side.] [Pythagorean Theorem (direct and inverse).]

In triangle $ABC$, the height $BM$, the bisector $BN$, and the median $BL$ are drawn. It is known that $AM = MN = NL$. Find the tangent of angle $A$ of this triangle.
Apply the property of the bisector of a triangle.

## Solution

Let $A M=M N=N L=x, L A=\alpha, A B=a$. Then

$$
C L=A L=3 x, A N=2 x, C N=4 x
$$

By the property of the bisector of a triangle

$$
\frac{A B}{B C}=\frac{A N}{N C}=\frac{1}{2}
$$

Therefore, $B C=2 A B=2 a$.

By the Pythagorean theorem

$$
B M^{2}=A B^{2}-A M^{2}=B C^{2}-C M^{2} \text {, or } a^{2}-x^{2}=4 a^{2}-25 x^{2} \text {. }
$$

From this, we find that $a=2 x \sqrt{2}$. Therefore,

$$
\cos \alpha=\frac{A M}{A B}=\frac{x}{a}=\frac{1}{2 \sqrt{2}}, \operatorname{tg} \alpha=\sqrt{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-05.jpg?height=400&width=778&top_left_y=2498&top_left_x=640)