In triangle $A B C$, the bisector $A D$ of angle $A$ and the bisector $B L$ of angle $B$ intersect at point $F$. The measure of angle $L F A$ is $60^{\circ}$.

1) Find the measure of angle $A C B$.
2) Calculate the area of triangle $A B C$, if $\angle C L D=45^{\circ}$ and $A B=2$.
Let $\angle A C B=\alpha$. Then

$$
\angle C A B+\angle C B A=180^{\circ}-\alpha, \angle L F A=\angle F A B+\angle F B A=\frac{1}{2}(\angle C A B+\angle C B A)=90^{\circ}-\frac{\alpha}{2}
$$

Since $\angle L F A=60^{\circ}$, then $\frac{\alpha}{2}=90^{\circ}-60^{\circ}=30^{\circ}$. Therefore, $\angle A C B=\alpha=60^{\circ}$.

Since $\angle L F D=180^{\circ}-\angle A F L=120^{\circ}$, a circle can be circumscribed around quadrilateral FLCD.

The inscribed angles $C F D$ and $C L D$ subtend the same arc, so $\angle C F D=\angle C L D=45^{\circ}$. Then $\angle$ $A F C=180^{\circ}-45^{\circ}=135^{\circ}$.

Applying the same reasoning used to calculate angle $A C B$, we find that $\angle A B C=90^{\circ}$. Thus, triangle $A B C$ is a right triangle, $B C=A B \cdot \operatorname{ctg} 60^{\circ}=\frac{2}{\sqrt{3}}$. Therefore,
![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-27.jpg?height=712&width=784&top_left_y=804&top_left_x=634)

## Answer

If $\angle A C B=\alpha$, then $\angle L F A=90^{\circ}-\frac{\alpha}{2}$.