Given any triangle $ABC$, let draw two circles with centers in $A$, $C$ and passing by $B$.
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 $H$, $G$.
My conjecture is that the points $AFCGH$ always determine a circle.
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!
Best Answer
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