Problem 4. Given is $\triangle A B C$ with the angle bisector of the angle at vertex $A$, $A L(L \in B C)$. The point $M \in A C$ is such that $M L \| A B$. Let $\beta^{\prime}$ be the exterior angle at vertex $B$ of $\triangle A B C$ and $\measuredangle A C B: \beta^{\prime}=5: 11$.

a) Determine the ratio $\measuredangle C A B$ : $\measuredangle A C B$.

b) If $\measuredangle A L B=80^{\circ}$, determine $\measuredangle A L M$.
Solution. a) According to the condition of the problem, we have $\measuredangle A C B=5 a$ and $\beta^{\prime}=11 a$, so $\measuredangle B A C+5 a=11 a$, hence $\measuredangle B A C=6 a$. Therefore, $\measuredangle C A B: \measuredangle A C B=6 a: 5 a=6: 5$.

![](https://cdn.mathpix.com/cropped/2024_06_05_8aa76aa53b232df96dffg-12.jpg?height=382&width=582&top_left_y=405&top_left_x=572)

b) From the condition of the problem it follows:

$\measuredangle B A L=\measuredangle L A M=$ For (the angle bisector at vertex $A$),

$\measuredangle B A L=\measuredangle A L M=3 a$ (pairs of alternate interior angles),

$\measuredangle A L B=\measuredangle L A C+\measuredangle A C L$ (an exterior angle for $\triangle A L C$).

Thus, $80^{\circ}=3 a+5 a$, so $a=10^{\circ}$ and $\measuredangle A L M=3 a=30^{\circ}$.