If $E$ is on the interior of $\triangle ABC$ this problem forces $ABC$ equilateral, from which $\angle BDC$ can vary a lot while $\angle CAE = 0$, so I don't see how this is true.

Yeah, of course angle conditions should be different, but I do not remember the conditions precisely

Thanks to an aops user, the condition has been changed

I really wanna see someone bash this problem in any coordinates... I got a synthetic solution: Let $w$ be the circle with center $B$ and radius $AB$. Let $L$ be the intersection of angle bisector of $\angle EBC$ and $w$. Let $\angle EBC=2\alpha$. Then $\angle BAE=60^o - 2\alpha$, then $\angle ABE=60^o + \alpha$, then $\angle BAC=60^o-1.5\alpha$, then $\angle EAC=0.5\alpha$ Now notice that $AE=AB=BL$ and $\angle EBL=\frac{\angle EBC}{2}=\alpha$, $\angle EAL=0.5\alpha+\frac{\angle LBC}{2}=\alpha$. Then by the Laws of Sines in triangles $AEL$ and $BEL$, $sin \angle ELA=sin \angle LEB$. Now $\angle BEL=360^o-(60^o+\alpha)-(180^o-\angle ELA -\alpha)=120^o+\angle ELA$. Now remember that their sines are equal. Obviously they are different, so they sum up to $180^o$, hence $\angle ELA = 30^o$. Then $ADEL$ is concyclic, and thus $\angle LDC=\angle LAE=\alpha=\angle LBC$, therefore $DBCL$ is also concyclic and $$\angle BDC + \angle CAE = \angle BLC + \angle CAE = 90^o-0.5\alpha+0.5\alpha=90^o$$