Question : Let $T$ be the centroid of a triangle $ABC$, and let $P$ be the midpoint of the side $\overline{AC}$. The line containing the point $T$ parallel to the line $BC$ intersects the side $\overline{AB}$ at the point $E$. Prove that the equality $\angle AEC = \angle PTC$ is true if and only if $\angle ACB=90^{\circ}$.
My attempt : 
$\angle EOT =\angle BOC, ET\left | \right | BC, \Rightarrow \angle TEO=\angle OCB. \Rightarrow \Delta EOT \sim \Delta OCB. \Rightarrow \frac{EO}{OC} =\frac{TO}{OB}. And \angle EOB=\angle TOC. \Rightarrow \Delta EBO \sim \Delta TOC. \Rightarrow \angle BEO=\angle OTC. \Rightarrow 180^{\circ}-\angle BEO=180^{\circ}-\angle OTC. \Rightarrow \angle AEC=\angle PTC.$
I have tried like this which is not requiring $\angle ACB=90^{\circ}$. But in the question we have to prove that $\angle AEC = \angle PTC$ is true if and only if $\angle ACB=90^{\circ}$. Can you please help me to prove it in the right way. So many thanks in advance.
$\endgroup$1 Answer
$\begingroup$Hint 1:
Angle condition: $\angle AEC = \angle PTC$ is equivalent to $\angle BEC = \angle BTC$ and this is equivalent to $C,B,E,T$ being concyclic.
Hint 2:
Deduce that $DA = DB =DC$.
Hint 2':
$\endgroup$2In reverse, deduce that $TC =EB$ so $CBET$ is isosceles trapezoid...
