In $\triangle ABC$, $AD$ $\perp$ $BC$ and $GE$ is the extended line of $DG$ where $G$ is centroid. Prove that $GD$ = $\frac{EG}{2}$ 
Let $ABC$ be a triangle and in $\triangle ABC$, $AD$ $\perp$ $BC$ and three median lines intersect at point $G$ where $G$ is the centroid of $\triangle ABC$. The extension of $DG$ intersects the circumcircle of $\triangle ABC$ at point $E$. Prove that $$GD = \frac{EG}{2}$$


I found this as an isolated problem.
My attempt:
Nothing speciality I discovered from the diagram. I only connected segment $AE$ and drew $GI$, where $GI$ $\perp$ $AD$.

From the above diagra, $G$ is the centroid. So, $\frac{AG}{GF}$ = $\frac{1}{2}$. And then from right angled triangle $\triangle AGI$ and $\triangle ADF$, We get $AI:ID$ = $1:2$ (as $\triangle AGI$ $\sim$ $\triangle ADF$). 
Right then, if $\triangle ADE$ can be showed as a right angled triangle ($\angle EAD$ = 90$^\circ$) and $\triangle ADE$ $\sim$ $\triangle GID$, we can also likewise show that $\frac{DG}{GE}$ = $\frac{1}{2}$. 
But my reverse effort went into vain. I can't anyhow show that $\angle EAD$ = 90$^\circ$. So, how to solve for that case?
SOURCE: BANGLADESH MATH OLYMPIAD
Can it be solved by vector? Thanks in advance.
 A: Before solving the problem, let's put ourselves in this situation (see attached image) 

If $HD=DJ, HG = 2(GO)$ and $JO=OI$ then you have to necessarily $J, O$ and $I$ are collinear.
A simple test is by contradiction. Suppose they are not collinear, so let's place $L$ in $HO$ such that $DL // JO$ then we would have $HL = 3a$, $LG = a$ and $GO = 2a$ ($a$ is a constant), in addition $DL = 2 (JO) = 2 (OI)$. With all this, we would have that $DL // OI$ (the triangles $DLG$ and $GOI$ are similar). Let's call $\theta = \angle JOL$, then $\angle DLH = \theta$ and $ \angle DLO = \angle LOI = 180 - \theta$, therefore $J,O,I$ are collinear. 
In the problem, let's call $H$ the orthocenter of the triangle $ABC$ and $O$ to its circumcenter. It is known that $H,G,O$ are collinear and $HG = 2GO$. Now let $J$ be the intersection of the extension of $AH$ with the circumscribed circumference, so it is easy to see that $HD = DJ$. Finally, we would have by the initial observation that necessarily J, O, I are collinear. Then $G$ is the centroid of the triangle $HIJ$ and $GI=2DG$.
Sorry for having changed the letter $E$ for the $I$, I hope it is understood in the same way
A: 
What you only have to prove is that $[CE]\parallel [AB]$. From there it's trivial (observe the ratio $CG:GF$) that $$[GD]=\frac{[EG]}{2}$$

So here goes my proof for $[CE]\parallel [AB]$ (I guess there might be a simpler one, however...)

Here $FK$ is the perpendicular bisector of $[AB]$, $K=DE\cap FK$ and $I$ is the midpoint of $[CD]$. Now since $CD\parallel KF$ $$\frac{[CD]}{[FK]}=\frac{[CG]}{[GF]}\implies [CD]=2·[FK]$$ Therefore $[KI]\parallel [AB]$, which implies that $[CK]=[DK]$. 
Denote by $O$ the circumcenter. 
Simple angle-chasing shows that $\angle CKO=\angle OKE$. From the congruence criterion SAS we obtain $$\Delta OKC\cong \Delta OEK\implies \angle KOC=\angle EOK$$
Thus the triangle $\Delta OEC$ is isosceles; the angle bisector of $\angle EOC$ is $OK\perp CE$. We can now conclude that $$CE\parallel AB$$ 
A: Let $E'$ be such that $ABCE'$ is isosceles trapezoid ($AB = CE'$).
Then $E'$ is on circle through $A,B,C$. Let $DE'$ meet $AF$ at $G'$.
Since triangle $DFG'\sim E'AG'$ and $AE' = 2DF$ we have $${AG'\over G'F} = {E'G'\over G'D} = {2\over 1}$$ so $G'=G $ and $E'=E$ and the claim is proved.
