<s>[INST] Prove that G is laying on K and find the ratio of AH to HB. We are given $Δ$ $ABC$ with $∠C = 90°$ and $AC > BC$. The circle K has a diameter CH (where CH is a height in $Δ$ $ABC$) and intersects with $AC$ in point $P$ and with $BC$ in point $Q$. If $PQ$ bisects $CG$, where $G$ is the centroid of $Δ$ $ABC$, prove that $G$ is laying on $K$ and find the ratio of $AH$ to $HB$.

You can find the diagram here
 [/INST] 
We have that $PQ$ is another diameter of the circle having $CH$ as a diameter and $CM\perp PQ$.
Assuming $BC=a<b=AC$, we have $CH=PQ=\frac{ab}{\sqrt{a^2+b^2}}$ and $$CG=\frac{2}{3}CM=\frac{2}{3}AM=\frac{\sqrt{a^2+b^2}}{3}.$$
Moreover, since $CPQ$ is similar to $CBA$,
$$ CP = \frac{a}{\sqrt{a^2+b^2}}PQ = \frac{a^2 b}{a^2+b^2},\qquad CQ=\frac{ab^2}{a^2+b^2}.$$
If $G$ lies on the circumcircle of $CPQ$, by Ptolemy's theorem
$$ CG\cdot PQ = 2\,CP\cdot CQ $$
has to hold, hence
$$ \frac{\sqrt{a^2+b^2}}{3}\cdot \frac{ab}{\sqrt{a^2+b^2}} = \frac{2a^3b^3}{(a^2+b^2)^2} $$
or
$$ (a^2+b^2)^2 = 6a^2 b^2 $$
that is a second degree equation in $r=\frac{b}{a}$, leading to:
$$ r=\frac{b}{a}=\color{red}{\sqrt{2+\sqrt{3}}}. $$
The remaining part is easy: $\frac{HA}{HB}=\frac{b^2}{a^2}=r^2=\color{red}{2+\sqrt{3}}.$
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