The triangle $A B C$ is isosceles with $A B=B C$. The bisector of the angle $\angle C A B$ intersects the side $B C$ at point $D$. The difference between the measures of two interior angles of the triangle $A B D$ is $40^{\circ}$. Find the possible values of the angle $\angle A C B$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-23.jpg?height=349&width=317&top_left_y=1745&top_left_x=949)
Solution

Let $x=\angle B A C=\angle A B C$. Thus, $\angle A C B=180^{\circ}-2 x$. Consequently, the internal angles of triangle $A B D$ are $\angle B A D=x / 2, \angle D B A=x$ and $\angle A D B=180^{\circ}-3 x / 2$. We will consider all cases for which two angles can differ by $40^{\circ}$:
i) If $x-x / 2=40^{\circ}$, then $x=80^{\circ}$ and thus $\angle A C B=20^{\circ}$.

ii) Since $x>0$, we cannot have $x / 2-x=40^{\circ}$.

iii) If $180^{\circ}-3 x / 2-x / 2=40^{\circ}$, then $x=70^{\circ}$ and $\angle A C B=40^{\circ}$.

iv) If $x / 2-\left(180^{\circ}-3 x / 2\right)=40^{\circ}$, we get $x=110^{\circ}$. This leads to a contradiction, as in this case $\angle A C B=180^{\circ}-2 x<0$.

v) If $\left(180^{\circ}-3 x / 2\right)-x=40^{\circ}$, then $x=56^{\circ}$ and $\angle A C B=68^{\circ}$.

vi) Finally, if $x-\left(180^{\circ}-3 x / 2\right)=40^{\circ}$, it follows that $x=88^{\circ}$ and $\angle A C B=4^{\circ}$.

Therefore, the possible values of $\angle A C B$ are $4^{\circ}, 20^{\circ}, 40^{\circ}$ and $68^{\circ}$.