The inscribed circle tangent BC, CA, AB at D, E and F in ABC. Let FG be diameter of the circle and let FD and EG intersect at H. Prove that CH||AB

Question

I have ben stuck on the last part of solving the problem for a while and don't Really know how to Prove it. My main approach was to Prove that the angles $HCB$ and $ABC$ are equal and I know that the triangles $CDE$, $AEF$ and $BDF$ are isocelese triangles. Please help but don't spoil the solution. Thank you

Answer 1

Since $F$ and $G$ are antipodal in the incircle, $FEH$ is a right triangle. The quadrilateral $AFIE$ has two opposite right angles and we have $EF=2r\cos\frac{A}{2}$. Since $\widehat{EFD}=\frac{\pi-C}{2}$, it follows that: $$ FH = 2r \frac{\cos\frac{A}{2}}{\sin\frac{C}{2}} $$ and since $BFD$ is an isosceles triangle the distance of $H$ from the $AB$ side is given by: $$ d(H,AB) = \frac{2r}{\sin\frac{C}{2}}\cos\frac{A}{2}\cos\frac{B}{2} $$ and $d(C,AB)=d(H,AB)$ follows from simple trigonometric manipulations.
This proves the claim $CH\parallel AB$.

Answer 2

$$\measuredangle DHG=90^{\circ}-\measuredangle DGH=90^{\circ}-\measuredangle DFE=\frac{1}{2}\measuredangle ACB,$$ which says that $CE=CD=CH$ and from here

$\measuredangle DHC=\measuredangle BFD$ and we are done!