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Given any triangle $ABC$, let draw two circles with centers in $A$, $C$ and passing by $B$. enter image description here

These circles determine a point $F$, which corresponds to the (other) intersection of the two circles.

Let now prolongate the sides $AB$ and $BC$ in such a way that these prolongations intersect the two circles in $H$, $G$.

enter image description here

My conjecture is that the points $AFCGH$ always determine a circle.

enter image description here

Is there an elementary proof of such conjecture?

This post is related to this one A conjecture related to a circle intrinsically bound to any triangle.

I apologize in case this is an obvious result. Thanks for your help!

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  • $\begingroup$ If the angle at B is right or acute, H and G are undefined. $\endgroup$ – Steve B Jul 6 '18 at 7:35
  • $\begingroup$ You discovered this on your own? Bravo! $(+1)\,\color{darkorange}{\bigstar}$ $\endgroup$ – Mr Pie Jul 6 '18 at 7:42
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    $\begingroup$ Yes.... with the help of Geogebra! : ) However, thanks! $\endgroup$ – user559615 Jul 6 '18 at 7:43
  • $\begingroup$ I looked at your other conjecture, too. You are very observant! $\stackrel{\bullet\,\bullet}{\smile}$ $\endgroup$ – Mr Pie Jul 6 '18 at 7:44
  • $\begingroup$ The proofs are way more beautiful than the conjectures! $\endgroup$ – user559615 Jul 6 '18 at 7:46
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I'll first show that $ACGH$ is inscribed:

Since $\triangle ABH$ and $\triangle BCG$ are isosceles and furthermore $\angle ABH=\angle CBG$ then they are similar which implies that $\angle BAH= \angle BCG$ hence the claim.

The isoscelessness implies that $\angle CGB=\angle CBG=\angle CAB+\angle ACB$

But now similarly you can show that $\triangle ACB$ and $\triangle ACF$ are congruent - they have $3$ correspondingly equal sides

This implies that $\angle CGB=\angle CBG=\angle CAB+\angle ACB=\angle CAF+\angle ACF=180-\angle AFC$

Hence $AFCG$ is inscribed and since it has $3$ common points with $ACGH$ then we get that

$AFCGH$ is inscribed. QED

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    $\begingroup$ Beautiful! Thanks a lot @asdf ! $\endgroup$ – user559615 Jul 6 '18 at 7:40
  • $\begingroup$ @asdf Perhaps I'm misunderstanding something. You seem to be saying that ∠BAH=∠BCG implies that A, C, G and H lie on a circle. Why is that? Certainly if the 4 points are on a circle those angles would be congruent, but what makes the converse true in this case? $\endgroup$ – Steve B Jul 6 '18 at 8:24
  • $\begingroup$ The result is if and only if. One can see that if you look at the locus of point which "see" $GH$ at a fixed angle - it is the disjoint union of $2$ arcs $\endgroup$ – asdf Jul 6 '18 at 8:27

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