# 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 $G$?

• Does it hold for all segments $\bar{EF}$? – Don Larynx Nov 14 '13 at 20:55
• yes I've checked it at geogebra – Marco Nov 14 '13 at 21:01
• Is $\Gamma$ specifically the incircle of $\triangle ABC$? – Blue Nov 14 '13 at 21:02
• @Blue: I don't think it is given, because $A$ and $C$ are not mentioned in the problem. I am thinking we should take $AC$ tangent to $\Gamma$ and parallel to $EF$ because I love similar triangles, but I haven't gotten there yet. – Ross Millikan Nov 14 '13 at 21:05
• @Blue: yes. I forgot but there is also given third point $K$ which lie at $\Gamma$ and it's tangent to $ABC$ – Marco Nov 14 '13 at 21:19

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. $\square$
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 $\Gamma$.