In a triangle $ABC$ if $AD$, $BE$, $CF$ are the angle bisectors of $\angle A$, $\angle B$ $\angle C$ respectively with incenter $I$ and $\angle A=120^\circ$ Find $\angle DFI$
It's $30^\circ$. If $BE$ meet $DF$ at $G$ and $CF$ meet $DE$ at $H$ it seem that $\triangle GEI$ and $\triangle HFI$ is an isosceles triangle. And also $GA$ and $HA$ is the angle bisector of $\angle BAD$ and $\angle CAD$ respectively. I dun know how to prove all of these, I just put it in GeoGebra and see the accurate picture, all I could find is $\angle BIC=150^\circ$, $\angle BIF=30^\circ$ and $\angle DFI+\angle DEI=60^\circ$. Please help
2 Answers
$\begingroup$In $\triangle BCD, \, $ $CA$ is the external angle bisector of $\angle DCH \, $ and $BE \,$ is the internal angle bisector of $\angle DBC$ and that makes point $E$ as the ex-center. Hence $DE$ is the external angle bisector of $\angle ADC$.
(As the bisectors of the external angles on one side of a triangle and the bisector of the opposite internal angle are concurrent and in this case the concurrency point is $E$).
say $\angle BEC = \alpha, \angle DEB = \beta$
Then $\angle CDE = 180^0 - (60^0 + \alpha + \beta) = 120^0 - \alpha -\beta \,$ (in $\triangle CED$)
Also $\angle ADE = \angle DEB + \angle DBE = \beta + \angle CBE = 60^0 - \alpha + \beta$
As $\angle CDE = \angle ADE, \angle DEB = \beta = 30^0$
By the way, in any triangle with $\angle C$ as $120^0$, the angle bisector of $\angle C$ meeting at point $D$ on $AB$ will make a right angle with the points where other two angle bisectors meet the other two sides of the triangle.
$\endgroup$ 2 $\begingroup$Construct an equilateral triangle $CFK$.
Since $\angle FAC + \angle FKC = 180^\circ$, $AFKC$ is a cyclic quadrilateral.
$\Rightarrow \angle FAK = \angle FCK = 60^\circ = \angle FAD$
$\Rightarrow AIDK$ is a straight line.
$\Rightarrow \angle DKC = \angle AFC = 180^\circ - \angle FAC - \angle ACF = 60^\circ - \frac{\angle C}{2} = \angle DCK$
$\Rightarrow DK = DC$
$\Rightarrow \triangle FDK \cong \triangle FDC$
$\Rightarrow \angle DFI = \frac{60 ^\circ}{2} = 30^\circ$
