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I am trying to solve this question in my own (Euclidean) way but was not successful. The question is interesting because the given is simple enough and therefore its solution should not be that difficult.

I think, it will be completely solved if we can prove that:-

KO extended (where O is the circum-center of ⊿ABC) meets AH extended right at the circum-circle.

Or equivalently, can we show that the mentioned intersection point is con-cyclic with A, B, and C.

I tried angle chasing and got the following:-

enter image description here

Alternate method that I have tried:-

enter image description here

Note that I am asking a different question though the given is the same.

I hope the proof need not go through showing that AK // BC.

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Mick
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  • @Futurologist This question cannot be considered as solved. Your posted answers are aiming at proving AK // BC. My post, though has the same set of given as #1978670, is asking for a different answer, namely “Will AH produced meet KO extended (where O is the circum-center of $\triangle ABC$) at the circum-circle?” In addition, I hope the proving process need not to go through the showing of AK // BC. – Mick Oct 28 '16 at 13:08
  • It is solved. Geometrically, in two different ways. See my edit below. – Futurologist Oct 28 '16 at 14:12
  • @CameronWilliams Noted. Pls see my comment to the following answer. – Mick Oct 28 '16 at 17:22

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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 point $H$. Since $ABC$ 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 $A$ so line $MH \equiv BC$ is mapped to line $AK$ so $BC$ is parallel to $AK$. As such, angle $\angle \, KAQ = 90^{\circ}$ and so $KQ$ is a diameter of the circumcircle of triangle $ABC$ which means that the center, point $O$, lies on $KQ$.

enter image description here

Futurologist
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  • Yes. I can see that the deduction is logical. However, you have done it in the order “If A, then C, and hence B.” I am wondering (my request) is it possible to have the thing done in the order of “If A, then B, and hence C”? – Mick Oct 28 '16 at 15:56
  • @Mick /yes, it can, but basically it is going to be like beating about the bush. And somewhere on the way one has to prove that some line is parallel to another... even if the two lines are not $AK$ and $BC$. – Futurologist Oct 28 '16 at 15:59