Another conjecture about a circle intrinsically bound to any triangle

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