In the following figure prove that: $AK||BC$ 
In arbitrary triangle $ABC$, let $G$ be centroid and $AH$ be a height.We extend $HG$ beyond $G$ to intersect circumcircle at $K$.Prove that $AK||BC$ 
  

I connected midpoint of $BC$ to $K$ and $A$ to show $MA=MK$ but failed...
 A: Let $M$ be the midpoint of edge $BC$ and let $O$ be the center of the circle $k$ circumscribed around triangle $ABC$. Then the line $OM$ is the orthogonal bisector of edge $BC$. 
Reflect triangle $ABC$ with respect to $OM$ and let the image of $A$ be point $A'$. Then triangle $ABC$ is mapped to $A'CB$ by the  reflection with respect to $OM$ so the two triangles are congruent. Because of that, $\angle \, BAC = \angle \, BA'C$ which implies that point $A'$ lies on the circle $k$. 

Let $H'$ be the image of $H$ with respect to the reflection in line $OM$. Then $A'H'$ is the image of the altitude $AH$ of triangle $ABC$ from vertex $A$ so $A'H'$ is the altitude of triangle $A'CB$ from the vertex $A'$. 
Since $AH = A'H'$ and $AH$ and $A'H'$ are both orthogonal to $BC$, the quad $AHH'A'$ is a rectangle and point $M$ is the midpoint of $HH'$. Therefore $HM = \frac{1}{2} HH' = \frac{1}{2} AA'$. 
On other hand, the centroid $G$ of triangle $ABC$ lies on the median $AM$ and $MG = \frac{1}{2} AG$. Therefore (by Thales' theorem or whatever the name of that rule is, or if you prefer by triangle similarity) the points $H, G, A'$ are collinear and $HG = \frac{1}{2} A'G$. However, by assumption, the ray $HG$ (the half-line starting from point $H$ through point $G$)  intersects the circle $k$ in a unique point already denoted by $K$. Therefore, $K\equiv A'$ and so $AK \equiv AA'$ is parallel to $BC$.          
Collinearity of $H, G, A'$. First method: The relations $$\frac{MH}{AA'} = \frac{MG}{AG} = \frac{1}{2}$$ and $\angle \, HMG = \angle \, HMA = \angle \, A'AM = \angle \, A'AG$, because $HM$ is parallel to $AA'$, imply that the two triangles $HMG$ and $A'AG$ are similar, so $\angle \, HGM = \angle \, AGA'$. Observe that the centroid point $G$ lies on the median $AM$ so $\angle \, AGA' + \angle \, MGA' = \angle \, AGM =  180^{\circ}$. But as $\angle \, AGA' = \angle \, HGM$ we have that $$\angle \, HGA' = \angle \, HGM + \angle \, MGA' = \angle \, AGA' + \angle \, MGA' = 180^{\circ}$$ which means that $H, G, A'$ are collinear.
Second method: Draw the line $HA'$ an let it intersect the median $AM$ at a point $G'$. Observe that $HM$ is parallel to $AA'$. Then by the intercept theorem (aka Thales' theorem or something like that)
$$\frac{MG'}{G'A} = \frac{HM}{AA'} = \frac{1}{2}$$ However, the centroid $G$ of triangle $ABC$ lies on the median $AM$ and splits it into
$$\frac{MG}{GA} = \frac{1}{2} = \frac{MG'}{G'A}$$  which is possible if and only if $G \equiv G'$. 
Short argument: Reflect in the line $OM$, which is the orthogonal bisector of $BC$. Since the circle $k$ is mapped to itself by the reflection, the image $A'$ of vertex $A$ lies on the circle and $AH$ is mapped to $A'H'$ with $M$ midpoint of $HH'$. Segment $AH$ is an altitude so $AHH'A'$ is a rectangle. Since $1:2 = HM : AA' = MG : GA$ by the intercept theorem $H,G,A'$ are collinear and so $K \equiv A'$.
A: Without loss of generality, we assume that the circum center is at the origin and that the circum radius is 1. Let $z_1, z_2, z_3$ be the complex numbers representing $A, B, C$ respectively. Then the centroid $G$ is $g = \dfrac{z_1+z_2+z_3}{3}$ and the point $H$ is $$\frac{1}{2}\left(z_1+z_2+z_3 - \frac{z_2z_3}{z_1}\right)$$ If $k$ is the complex number representing $K$, then equating the slopes of $KG$ and $GH$, we have 
\begin{align*}
\frac{g - k}{\bar{g}-\bar{k}} &= \frac{g - \frac{1}{2}\left(z_1+z_2+z_3 - \frac{z_2z_3}{z_1}\right)}{\bar{g} - \frac{1}{2}\left(\bar{z_1}+\bar{z_2}+\bar{z_3} - \frac{\bar{z_2}\bar{z_3}}{\bar{z_1}}\right)}\\
&= \frac{g - \frac{z_2z_3}{z_1}} {\bar{g}-\frac{\bar{z_2}\bar{z_3}}{\bar{z_1}}}
\end{align*}
Thus the line joining $G$ and $K$ and that joining $G$ and $\frac{z_2z_3}{z_1}$ have the same slope. If $AH$ extended meets the circle again in $A'$, then the complex number representing $A'$ is $-\frac{z_2z_3}{z_1}$ and hence the point $\frac{z_2z_3}{z_1}$ is the point diametrically opposite to $A'$. Thus $K$ is the point $\frac{z_2z_3}{z_1}$. Now, slope of $AK$ is $-z_1\frac{z_2z_3}{z_1} = -z_2z_3$ and equals the slope off $BC$. Thus $AK \parallel BC$.
A: Let $L, M, N$ be the midpoints of edges $AB, BC, CA$ respectively. Then the circumcircle of triangle $LMN$ is the six point circle and passes through the feet of the three altitudes of triangle $ABC$. In particular it passes through the point $H$. Alaternatively, you can see that by either arguing that triangles $LHN$ and $LMN$ are congruent, or by noticing that $\angle \, LMN = \angle \, LAN = \angle \, LHN$, which means that $H$ lies on the circle circumscribed around triangle $LMN$. 
Since triangle $ABC$ is the homothetic image of $LMN$ with respect to the homothety with center the centroid $G$ and scaling factor $-2$, the six point circle is mapped to the circumcircle of $ABC$. Therefore, point $H$ is mapped to point $K$ and point $M$ is mapped to point $A$. Consequently, line $MH \equiv BC$ is mapped to line $AK$ so $BC$ is parallel to $AK$.

A: Equivalent problem:
Let $AK||BC$, $M$ the midpoint of $BC$ and $HK$ intersects $AM$ at $G$. I'll show that $G$ is the centroid of $ABC$.
I draw $KE \bot BC$, and because $AKCB$ is isosceles trapezoid, $M$ is the midpoint of $HE$. Hence: 
$\frac{{AG}}{{GM}} = \frac{{AK}}{{HM}} = 2$ , so $G$ is centroid.

