Let D be an interior point of the side BC of a triangle ABC . Let...

Let D be an interior point of the side BC of a triangle ABC . Let

\mathrm{I}_1

and

\mathrm{I}_2

be the incentres of triangles

A B D

and

A C D

respectively. Let

A I_1

and

A I_2

meet

B C

E

and

F

respectively. If

\angle B I_1 E=60^{\circ}

. What is the measure of

\angle C I_2 F

in degrees? Let D be an interior point of the side BC of a triangle ABC . Let I_1 and I_2 be the incentres of triangles A B D and A

Correct answer

\begin{aligned} & \angle \mathrm{BAD}=120^{\circ}-\mathrm{B} \\ & \angle \mathrm{CAD}=\angle \mathrm{A}-\left(120^{\circ}-\mathrm{B}\right) \\ & =\mathrm{A}+\mathrm{B}-120^{\circ} \\ & \angle \mathrm{FAC}=\frac{\mathrm{A}+\mathrm{B}}{2}-60^{\circ} \Rightarrow 90^{\circ}-\frac{\mathrm{C}}{2}-60^{\circ} \Rightarrow 30^{\circ}-\frac{\mathrm{C}}{2} \\ & \angle \mathrm{AFC}=180^{\circ}-\left(\mathrm{C}+30^{\circ}-\frac{\mathrm{C}}{2}\right) \\ & =150-\mathrm{C} / 2 \\ & \angle \mathrm{Cl}_2 \mathrm{~F}=180^{\circ} \mathrm{m}-\left(150^{\circ}-\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{2}\right)=30^{\circ}\end{aligned}