prove that center of circle lie on the other circle There is a $\triangle ABC$ , $D$ is the center of the circle $\Gamma$, tangent to the triangle at points $E, F$. How to prove that center of the circle inscribed in a $\triangle BEF$ lies at circle $\Gamma$ ? I know that $\triangle BEF$ is isosceles hence $H$ is the middle of $EF$, but how to prove that bisector of $\angle BFE$ or $\angle BEF$ intersects with that of $\angle FBE$ at $G$?
 
 A: We can proceed by considering only elements "below" $\overline{BD}$.



*

*Right triangles $\triangle EHD$ and $\triangle BED$ share an angle at $D$, so that $\angle DEH \cong \angle DBE$.

*$\overline{EG}$ bisects $\angle HEB$ (as $G$ is the intersection of angle bisectors in $\triangle BEF$), so that $\angle HEG \cong \angle BEG$.

*$\angle EGD$ is an external angle for $\triangle EGB$, so that its measure is the sum of those of the remote interior angles $$\angle GED = \angle BEG + \angle DBE = \angle HEG + DEH = \angle DEG$$

*Therefore, $\triangle DEG$ is isosceles with base $\overline{GE}$. The other two sides are congruent radii of circle $\Gamma$; in particular $\overline{DG}$ is a radius. $\square$




A: Notice that $\overline{DF}\perp \overline{BF}$ and $\overline{DE}\perp \overline{BE}$, so $\angle DFB$ and $\angle DEB$ are supplementary.  Thus, $DEBF$ must be a cyclic quadrilateral.  It follows that $\angle BFE=\angle BDE$.  Since the center of the incircle of a triangle lies on the intersection of the triangle's angle bisectors and $G$ is the incenter of $\triangle BEF$, $2\angle GFE=\angle BFE=\angle BDE$.  As the angle from the center of a circle is twice the angle subtended by the same arc, $\overline{GF}$ must intersect the circle at the same point $\overline{BD}$ does.  However, we also know that $G$ must lie on $\overline{BD}$, since $\overline{BD}$ is the angle bisector of $\angle ABC$.  Thus, $G$ is the intersection of $\overline{BD}$ with circle $\Gamma$.
