New problem

[NaziaC]

In triangle $ABC$, $G$ is the centroid and $D$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $E$. Prove that \[\angle AEC= \angle DGC\] if and only if \[\angle ACB= 90^{\circ}\].

[Corei13]

HINT:

$\angle AEC = \angle AEG + \angle GEC$
$= \angle EBG + \angle GBC + \angle ECB$
and, $\angle DGC = \angle GCB + \angle GBC$
$= \angle GCE + \angle ECB + \angle GBC$

$ \Longrightarrow \angle EBC = \angle GCE $
$ \Longrightarrow \angle GCB = \angle B$

[Masum]

Only if part:
Let $AH,BD,CF$ be medians,$C=90,GE $ intersect $AB$ at $L,BG$ intersect $CE$ at $K$.
$F$ is the circumcentre of $ABC$,so $FB=FC,FG=FE$.Because $FH$ is the perpendicular bisector of $GE,\Delta GKE$ is isosceles.
So $\angle AEC=\angle AEG+\angle KEG=\angle KGE+\angle FGE=\angle DGL+\angle CGL=\angle DGC$