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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 $HB$.

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You can find the diagram here

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  • $\begingroup$ Your diagram is not fully visible $\endgroup$ – Jean Marie Sep 21 '16 at 18:41
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    $\begingroup$ A suggestion to improve readability: export your drawing with Geogebra as a png file and include it in your post directly. $\endgroup$ – Jack D'Aurizio Sep 21 '16 at 18:42
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    $\begingroup$ Is $H$ the foot of the altitude from $C$ ? $\endgroup$ – Jean Marie Sep 21 '16 at 18:43
  • $\begingroup$ Yes, H is the foot of the altitude from C. $\endgroup$ – Gale Staneva Sep 21 '16 at 18:48
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    $\begingroup$ In the problem CH was an altitude but I missed to write it in the problem's description :| $\endgroup$ – Gale Staneva Sep 21 '16 at 20:28
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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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  • $\begingroup$ Thank you! I just discovered that we can also use the power of the point M for the last part but your solution is appealing $\endgroup$ – Gale Staneva Sep 21 '16 at 19:51
  • $\begingroup$ Well, how do you know $CPQ$ is similar to $ABC$? The problem doesn't state the $CH$ is the altitude. You have to prove it. The rest is easy. $\endgroup$ – Futurologist Sep 21 '16 at 20:20
  • $\begingroup$ @Futurologist: the problem states that $CH$ is an altitude (see OP's comment) and the similarity follows from $CM\perp PQ$. $\endgroup$ – Jack D'Aurizio Sep 21 '16 at 20:22
  • $\begingroup$ Otherwise, you have to prove that $CH$ is an altitude starting from what? $\endgroup$ – Jack D'Aurizio Sep 21 '16 at 20:22
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    $\begingroup$ @JackD'Aurizio yes, the two solutions together make the final solution. Agreed. $\endgroup$ – Futurologist Sep 21 '16 at 20:28
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$CH$ is not assumed to be the altitude. One has to prove it. It follows from the assumptions of the problem. Here is how you can prove it.

If $PQ$ bisects orthogonally $CG$ then $G$ is the symmetric image of $C$ with respect to $PQ$ and triangles $PCQ$ and $PGQ$ are congruent. Hence $\angle \, PGQ = 90^{\circ} = \angle \, PCQ$ and since $\angle \, PGQ + \angle \, PCQ = 180^{\circ}$ the quadrilateral $QCPG$ is inscribed in $K$. Since $CH$ is a diameter, the angles $\angle \, HQC = \angle \, HPC = 90^{\circ}$. Moreover, $\angle \, PCQ = 90^{\circ}$ so $PCQH$ is a rectangle inscribed in $K$ and the center of the triangle is the centroid of the rectangle. Hence, triangles $PHQ, \, PCQ$ and $PGQ$ are congruent. Segment $CE$ is a median in a right-angled triangle so $CE = BE = AE$ and the triangle $BCE$ is isosceles with $\angle \, EBC = \angle \, ECB = \angle \, GCQ$. By symmetry, $\angle \, GCQ = \angle \, CGQ$, and then $\angle \, CGQ = \angle \, CPQ$ as inscribed in $K$. We arrive at the conclusion that $\angle \, ABC = \angle \, EBC = \angle \, GCQ = \angle \, CPQ$. Therefore triangles $ABC$ and $QPC$ are similar and since $CH$ is a diameter of the circle $K$ circumscribed around $QPC$, it is it's median while $CE$ is the median of $ABC$ leading to the conclusion that $\angle \, BCH = \angle \, ACE = \angle EAC = \angle \, BAC$. But this means that $\angle \, HBC + \angle \, BCH = \angle \, ABC + \angle \, BAC = 90^{\circ}$ which means that $CH$ is orthogonal to $AB$.

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