Altitude Point, Centroid and Circumcircle Point are Collinear

Question

Let $ABC$ be an acute triangle and suppose $X$ is a point on the circumcircle of $ABC$ with $AX\parallel BC$ and $X\neq A$. Denote $G$ the centroid of triangle $ABC$, and by $K$ the foot of the altitude from $A$ to $BC$. Prove that $K,G,X$ are collinear.

I have tried angle chasing but it's hard to catch angles here

Answer 1

Let $AG$ meet $BC$ at $E$, and let $T$ be the point on $BC$ such that $XT\perp BC$. Because $AX//BC$, and $AK\perp BC$, we know $AXTK$ is a rectangle. Further, $E$ is the midpoint of $KT$, because $ABCX$ is an isosceles trapezoid. Therefore, we know $KE = \frac{1}{2}\cdot KT = \frac{1}{2}\cdot AX$, where $KT=AX$ because $AXTK$ is a rectangle.

Now, extend $KG$ to meet $AX$ at $X'$. Since $G$ is the centroid and $BC//AX$, we have $$\frac{KE}{AX'} = \frac{EG}{GA} = \frac{1}{2}\implies KE = \frac{1}{2}\cdot AX'$$ Hence, $AX=AX' \implies X=X' \implies K, G, X$ are on the same line (because we created $X'$ from extending $EG$).

Note: Proof that $ABCX$ is a isosceles trapezoid is based on $\angle ABC=\angle CAX = \angle BCA$, $\angle BAC = \angle BXC$ and $BC = CB$, so $\triangle ABC$ and $\triangle BXC$ are congruent, so $AB = XC$. Proof that $E$ is the midpoint of $KT$ is based on $\triangle ABK$ and $\triangle XCT$ being congruent, so $BK = CT$. Since $E$ is the midpoint of $BC$, we have $EK = BE-BK=CE-CT = ET$.

Answer 2

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 $K$. Alaternatively, you can see that by either arguing that triangles $LKN$ and $LMN$ are congruent, or by noticing that $\angle \, LMN = \angle \, LAN = \angle \, LKN$, which means that $K$ 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 $ABC$ and line $MK$, which coincides with $BC$, is mapped to the line $AX$ parallel to $BC$. Thus, point $X$ is the homothetic image of $K$ and therefore $K, G$ and $X$ are collinear.