Here is the whole problem:
In the triangle $ABC$, it is known that $AC > AB$, and the angle at the vertex $A$ is equal to $\alpha$. On the side $AC$, point $M$ is marked so that $AB=MC$. Point $E$ is the midpoint of the segment $AM$, point $D$ is the midpoint of the segment $BC$. Find the angle $CED$.
The textbook answer to the problem is that $\angle CED = \dfrac{\alpha}{2}$. The tip on how get to it is to mark a point $K$ on the $AB$ such, that $AK = KB$, draw the midline $KD$ of the triangle $ABC$ and then prove that the $KDE$ triangle is isosceles.
At that point I've tried everything(even chapgpt, which, as it turned out, is bad at math), the best I've gotten so far is if I mark a midpoint F on the $CM$, then I get parallelogram $EKDF$ $\bigg(KD = \dfrac{1}{2}AC = EF$, and $KD\parallel AC\bigg)$. And triangles $KDE$ and $FED$ are congruent, but it's not even close to what I need to prove. How would you do that ?
Here is the final figure:
Please, take into account that this is a problem from an $\mathbf{8^{th}}$ grade math textbook, the topic is "Midline of a triangle", which means I'm not allowed to use any angle functions or similar features that were not introduced yet in the curicullumn.
EDIT: The textbook's hint seems to be wrong, thanks @Vasili for the solution. The actual triangle that needs to be proven to be isosceles is another one, not the $KDE$.
Here is how I arrived to the solution but also without completely using the hint.
Draw $BM$ and $NE$. $NE$ is a midline in $\triangle ABM$ so $AKNE$ is a paralellogram, $NE=AK$. $ND$ is a midline in $\triangle BMC$ so $ND=\frac{MC}{2}=AK=NE \implies \triangle END$ is isosceles.