8.4. In triangle $A B C$, the bisectors $A D$ and $C E$ were drawn. It turned out that $A E + C D = A C$. Find the angle $B$.

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Answer: angle $B$ is equal to $60^{\circ}$.

Let $\alpha$ and $\gamma$ be the measures of angles $A$ and $C$ of triangle $ABC$. On side $AC$, we mark segment $AF$ equal to $AE$. From the condition, we get the equality of segments $CD$ and $CF$. Let the bisectors $AD$ and $CE$ intersect segments $EF$ and $FD$

![](https://cdn.mathpix.com/cropped/2024_05_06_24de5779eb3996ecd66eg-2.jpg?height=394&width=460&top_left_y=268&top_left_x=1415)
at points $K$ and $L$. In the isosceles triangle $FAE$, segment $AK$ is a bisector, hence a median and an altitude, as is segment $CL$ in $FCD$. In triangle $EFD$, segments $EL$ and $DK$ are medians and altitudes, so it is equilateral. Further, $\angle AFE = 90^{\circ} - \alpha / 2$, $\angle EFD = 60^{\circ}$, and $\angle DFC = 90^{\circ} - \gamma / 2$. Adding these angles and equating to $180^{\circ}$, we get $\alpha + \gamma = 120^{\circ}$. Therefore, angle $B$ is equal to $60^{\circ}$.