1
$\begingroup$

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 $G$?

$\endgroup$
  • 1
    $\begingroup$ Does it hold for all segments $\bar{EF}$? $\endgroup$ – Don Larynx Nov 14 '13 at 20:55
  • $\begingroup$ yes I've checked it at geogebra $\endgroup$ – Marco Nov 14 '13 at 21:01
  • $\begingroup$ Is $\Gamma$ specifically the incircle of $\triangle ABC$? $\endgroup$ – Blue Nov 14 '13 at 21:02
  • $\begingroup$ @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. $\endgroup$ – Ross Millikan Nov 14 '13 at 21:05
  • $\begingroup$ @Blue: yes. I forgot but there is also given third point $K$ which lie at $\Gamma$ and it's tangent to $ABC$ $\endgroup$ – Marco Nov 14 '13 at 21:19
1
$\begingroup$

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. $\square$

enter image description here

$\endgroup$
0
$\begingroup$

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 $\Gamma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.