4.22. In a right-angled triangle, find the bisector of the right angle, if the hypotenuse of the triangle is equal to $c$, and one of the acute angles is equal to $\alpha$.
4.22. In $\triangle A B C$ (Fig. 4.19), we have $A C=c \sin \alpha$, $B C=c \cos \alpha, B D=x, A D=c-x, l$ - the angle bisector of angle $C$. Since $\frac{x}{c-x}=\frac{c \cos \alpha}{c \sin \alpha}=\operatorname{ctg} \alpha$, then $x=$ $=c \operatorname{ctg} \alpha-x \operatorname{ctg} \alpha$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-582.jpg?height=383&width=248&top_left_y=1462&top_left_x=891)

Fig. 4.19

$$
x=\frac{c \operatorname{ctg} \alpha}{1+\operatorname{ctg} \alpha}=\frac{c \cos \alpha}{\sin \alpha+\cos \alpha}, c-x=\frac{c \sin \alpha}{\sin \alpha+\cos \alpha}
$$

Now we use the formula $l=\sqrt{A C \cdot B C-A D \cdot D B}$ and get

$$
\begin{aligned}
& l=\sqrt{c^{2} \sin \alpha \cos \alpha-\frac{c^{2} \sin \alpha \cos \alpha}{(\sin \alpha+\cos \alpha)^{2}}}=c \sqrt{\sin \alpha \cos \alpha \frac{(\sin \alpha+\cos \alpha)^{2}-1}{(\sin \alpha+\cos \alpha)^{2}}}= \\
& =c \sqrt{\frac{\sin 2 \alpha}{2} \cdot \frac{\sin 2 \alpha}{1+\sin 2 \alpha}}=\frac{c \sin 2 \alpha}{\sqrt{2\left(1+\cos \left(90^{\circ}-2 \alpha\right)\right)}}=\frac{c \sin 2 \alpha}{\sqrt{2 \cdot 2 \cos ^{2}\left(45^{\circ}-\alpha\right)}}= \\
& =\frac{c \sin 2 \alpha}{2 \cos \left(45^{\circ}-\alpha\right)}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-583.jpg?height=294&width=402&top_left_y=876&top_left_x=59)

Fig. 4.20

$$
\text { Answer: } \frac{c \sin 2 \alpha}{2 \cos \left(45^{\circ}-\alpha\right)}
$$

4. 23. I method. Let $B K=x, A L=$ $=2 y$ (Fig. 4.20), then by the condition $K C=$ $=3 x, L C=5 y$. Let, further, $K F=m$, $O F=n$. From the similarity of triangles $O K F$

$$
\text { and } A K C \text { we have } \frac{n}{m}=\frac{7 y}{3 x}, \text { from which }
$$

$$
\frac{n}{y}=\frac{7 m}{3 x}
$$

From the similarity of triangles $B O F$ and $B L C$ we find $\frac{n}{5 y}=\frac{x+m}{4 x}$, from which

$$
\frac{n}{y}=\frac{5(x+m)}{4 x}
$$

By equating the right-hand sides of proportions (1) and (2), after simplifications we get $28 m=15(x+m)$, from which $m=\frac{15 x}{13}$. Therefore,

$$
B F=B K+K F=\frac{28 x}{13} \text { and } C F=B C-B F=4 x-\frac{28 x}{13}=\frac{24 x}{13} \text {. }
$$

According to Thales' theorem, we have

$$
\frac{B O}{O L}=\frac{B F}{F C}=\frac{\frac{28 x}{13}}{\frac{24 x}{13}}=\frac{7}{6}, \frac{A O}{O K}=\frac{\frac{24 x}{13}}{\frac{15 x}{13}}=\frac{8}{5}
$$

II method. Let the sides of triangle $A B C$ represent weightless rods, and forces are applied at the vertices of the triangle (Fig. 4.21). Suppose that a force of $2 \mathrm{H}$ is applied at vertex $C$; then at point $B$ according to the equilibrium conditions (equality of moments of forces relative to point $K$) a force of $6 \mathrm{H}$ should be applied, and at point $K$ according to the rule of adding parallel forces a force of $8 \mathrm{H}$ should be applied. Reasoning similarly for points $A$ and $L$, we find that a force of $5 \mathrm{H}$ should be applied at point $A$, and a force of $7 \mathrm{H}$ at point $L$. Finally, according to the equilibrium conditions relative to point $O$ we get $A O: O K=8: 5$ and $B O: O L=7: 6$.

Answer: $8: 5$ and $7: 6$.

4. 24. Hint. For each of the medians of the triangle, use formula (1.35).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-584.jpg?height=477&width=469&top_left_y=642&top_left_x=685)

Fig. 4.21