12.143. The tangent of the angle at the base of an isosceles triangle is 3/4. Find the tangent of the angle between the median and the bisector drawn to the lateral side.
## Solution.

In $\triangle ABC$, $AB=BC$, $AM$ is the angle bisector, $AN$ is the median, $BD$ is the altitude, (Fig.12.16). $\angle C=\alpha, \operatorname{tg} \alpha=\frac{3}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0903.jpg?height=365&width=724&top_left_y=-1&top_left_x=292)

Fig. 12.16

Let $BD=3x \Rightarrow CD=BD \operatorname{ctg} \alpha=4x$.

Draw $NK \| BD$. Since $BN=NC$, then $DK=KC=2x, NK=\frac{1}{2} BD=1.5x$, $AK=AD+DK=6x \quad \angle MAC=\frac{\alpha}{2}$.

Let $\angle MAN=\gamma$, then $\angle NAC=\frac{\alpha}{2}-\gamma, \operatorname{tg}\left(\frac{\alpha}{2}-\gamma\right)=\frac{NK}{AK}=\frac{1.5x}{6x}=\frac{1}{4}$. $\operatorname{tg} \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{3}{4} ; 3 \operatorname{tg}^{2} \frac{\alpha}{2}+8 \operatorname{tg} \frac{\alpha}{2}-3=0 ; \Rightarrow \operatorname{tg} \frac{\alpha}{2}=\frac{1}{3} ;$

$\operatorname{tg} \frac{\alpha}{2}=-3$

Since $\alpha$ is an acute angle, $\operatorname{tg} \frac{\alpha}{2}>0, \operatorname{tg} \frac{\alpha}{2}=\frac{1}{3}$. Thus, $\operatorname{tg}\left(\frac{\alpha}{2}-\gamma\right)=\frac{\operatorname{tg} \frac{\alpha}{2}-\operatorname{tg} \gamma}{1+\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \gamma}=\frac{1}{4} ; \frac{\frac{1}{3}-\operatorname{tg} \gamma}{1+\frac{1}{3} \operatorname{tg} \gamma}=\frac{1}{4} ; \operatorname{tg} \gamma=\frac{1}{13}$.

Answer: $\frac{1}{13}$.