9.23*. (Jury, Netherlands, 79). Inside an equilateral triangle $ABC$, points $K, L, M$ are taken such that $\angle KAB = \angle LBA = 15^{\circ}, \quad \angle MBC = \angle KCB = 20^{\circ}$, $\angle LCA = \angle MAC = 25^{\circ}$. Find the angles of triangle $KLM$.[^2]

## § 10. Circles and Circles

(see Appendix G: definitions 35, 36; theorems $66-75,79$ )
9.23. Let $\alpha=20^{\circ}, \beta=25^{\circ}, \gamma=15^{\circ}$ (Fig. 19). Then $\alpha, \beta, \gamma<30^{\circ}, \alpha+\beta+\gamma=60^{\circ}, \angle K A M=60^{\circ}-\angle M A C-\angle K A B=\alpha$ and similarly $\angle L B M=\beta, \angle K C L=\gamma$. Let segments $A M$ and $C L$ intersect at point $N$, lines $B N$ and $A C$ at point $P$, and lines $C L$ and $A B$ at point $Q$. Since $A B=B C$ and $A N=N C$ (since $\angle N A C=$ $=\angle N C A=\beta$), $B P$ is the bisector of angle $A N C$, and thus of angle $L N M$. Let's take a point $B^{\prime}$ inside angle $L N M$ and outside triangle $L M N$, equidistant from the three lines $L N, L M$, and $M N$. This point lies on line $N B$ and on the bisectors of the external angles at $L$ and $M$ of triangle $L M N$, so

$$
\begin{aligned}
& \angle L B^{\prime} M=180^{\circ}-\angle B^{\prime} L M-\angle B^{\prime} M L= \\
& =180^{\circ}-\frac{180^{\circ}-\angle N L M}{2}-\frac{180^{\circ}-\angle N M L}{2}=\frac{\angle N L M+\angle N M L}{2}= \\
& =\frac{180^{\circ}-\angle L N M}{2}=\frac{\angle N A C+\angle N C A}{2}=\beta=\angle L B M
\end{aligned}
$$

Therefore, $B^{\prime}=B$ and

$$
\begin{aligned}
\angle B L M=\angle & B L Q=180^{\circ}-\angle C Q B-\angle A B L= \\
& =\angle A B C+\angle B C Q-\angle A B L=60^{\circ}+\alpha+\gamma-\gamma=60^{\circ}+\alpha
\end{aligned}
$$

Since

$\angle B L C=180^{\circ}-\angle L B C-\angle L C B=180^{\circ}-(\beta+\alpha)-(\alpha+\gamma)=120^{\circ}-\alpha$, then

$$
\angle M L C=\angle B L C-\angle B L M=120^{\circ}-\alpha-\left(60^{\circ}+\alpha\right)=60^{\circ}-2 \alpha
$$

(recall that $\alpha<30^{\circ}$). Similarly, it can be shown that

$$
\angle K L B=60^{\circ}-2 \alpha
$$

Thus, we have

$$
\begin{aligned}
\angle K L M=\angle B L C-\angle K L B- & \angle M L C= \\
& =\left(120^{\circ}-\alpha\right)-\left(60^{\circ}-2 \alpha\right)-\left(60^{\circ}-2 \alpha\right)=3 \alpha
\end{aligned}
$$

i.e.

$$
\angle K L M=3 \alpha=60^{\circ}
$$

Similarly, it can be shown that

$$
\angle L K M=3 \beta=75^{\circ} \text { and } \angle K M L=3 \gamma=45^{\circ}
$$

## § 10. Circles and Circles