Divided incenters

Geometry Level 3

Let D D be an interior point of the side B C BC of a triangle A B C ABC . Let I 1 I_1 and I 2 I_2 be the incentres of triangles A B D ABD and A C D ACD respectively. Let A I 1 AI_1 and A I 2 AI_2 meet B C BC in E E and F F respectively. If ∠ B I 1 E = 6 0 o \angle BI_1E = 60^o , what is the measure of ∠ C I 2 F \angle CI_2F in degrees?


The answer is 30.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Following the labelling of the diagram, we have

∠ B I 1 E = B 2 + θ ⇒ 60 ∘ = B 2 + θ ( 1 ) \angle B{{I}_{1}}E=\frac{B}{2}+\theta \Rightarrow 60{}^\circ =\frac{B}{2}+\theta \ \ \ \ \ (1)

∠ C I 2 F = C 2 + φ ( 2 ) \angle C{{I}_{2}}F=\frac{C}{2}+\varphi \ \ \ \ \ (2)

( 1 ) , ( 2 ) ⇒ + ∠ C I 2 F + 60 ∘ = ( C 2 + φ ) + ( B 2 + θ ) = B 2 + C 2 + ( θ + φ ) = B 2 + C 2 + A 2 = 90 ∘ \begin{aligned} \left( 1 \right),\left( 2 \right)\overset{+}{\mathop{\Rightarrow }}\,\angle C{{I}_{2}}F+60{}^\circ & =\left( \frac{C}{2}+\varphi \right)+\left( \frac{B}{2}+\theta \right) \\ & =\frac{B}{2}+\frac{C}{2}+\left( \theta +\varphi \right) \\ & =\frac{B}{2}+\frac{C}{2}+\frac{A}{2} \\ & =90{}^\circ \\ \end{aligned} Hence,

∠ C I 2 F = 90 ∘ − 60 ∘ = 30 ∘ \angle C{{I}_{2}}F=90{}^\circ -60{}^\circ =\boxed{30{}^\circ} .

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