Solution

Let's mark the points of intersection and the angles of interest on the figure, with $x$ being the angle of the intersection of the angle bisectors we are looking for.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-13.jpg?height=661&width=619&top_left_y=463&top_left_x=798)

By the sum of the interior angles of triangle $\triangle A B C$, we have:

$$
\begin{aligned}
100^{\circ}+\left(180^{\circ}-\alpha-3 \beta\right)+\left(180^{\circ}-\beta-3 \alpha\right) & =180^{\circ} \\
460^{\circ}-4 \alpha-4 \beta & =180^{\circ} \\
4 \alpha+4 \beta & =280^{\circ} \\
\alpha+\beta & =70^{\circ}
\end{aligned}
$$

From triangles $\triangle A B D$ and $\triangle B C E$, we have $\alpha+2 \theta=100^{\circ}$ and $\beta+2 \lambda=100^{\circ}$. Adding these two equations, we get:

$$
\begin{aligned}
\alpha+2 \theta+\beta+2 \lambda & =100^{\circ}+100^{\circ} \\
\alpha+\beta+2 \theta+2 \lambda & =200^{\circ} \\
70^{\circ}+2(\theta+\lambda) & =200^{\circ} \\
\theta+\lambda & =65^{\circ}
\end{aligned}
$$

Now, let's use the sum of the interior angles of quadrilateral $K B L J$.

$$
\begin{aligned}
\angle K J L+\angle B K J+\angle K B L+\angle B L J & =360^{\circ} \\
x+\left(80^{\circ}+\theta\right)+100^{\circ}+\left(80^{\circ}+\lambda\right) & =360^{\circ} \\
x+260^{\circ}+\theta+\lambda & =360^{\circ} \\
x+\theta+\lambda & =100^{\circ} \\
x+65^{\circ} & =100^{\circ} \\
x & =35^{\circ}
\end{aligned}
$$

#
In the figure, $\angle A B C=100^{\circ}, \angle F A C=3 \angle E C B$ and $\angle G C A=3 \angle D A B$. Determine the measure of the acute angle at the intersection of the internal angle bisectors of triangles $\triangle A D B$ and $\triangle C E B$ relative to angles $D$ and $E$.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-12.jpg?height=698&width=637&top_left_y=1642&top_left_x=591)