Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let ABCABC be a scalene triangle inscribed in circle Γ\Gamma. The internal bisector of ∠A\angle A meets BC‾\overline{BC} and circle Γ\Gamma at points DD and EE. The circle with diameter DE‾\overline{DE} meets Γ\Gamma at a second point FF. Prove that ABAC=FBFC\frac{AB}{AC}=\frac{FB}{FC}.

Solution

Solution:
Let KK be the midpoint of arc⁡BC^\operatorname{arc} \widehat{BC} containing AA. Let MM be the midpoint of BC‾\overline{BC}.
Figure 1
Since ∠DME=∠DFE=90∘\angle DME=\angle DFE=90^\circ, the points D,M,E,FD, M, E, F are concyclic. Moreover, since ∠EFK=∠EFD=90∘\angle EFK=\angle EFD=90^\circ, the points F,D,KF, D, K are collinear. Finally, since ∠KAD=∠KMD=90∘\angle KAD=\angle KMD=90^\circ, the points A,D,M,KA, D, M, K are concyclic.
Therefore,
∠FAD=∠FAE=∠FKE=∠DKM=∠DAM. \angle FAD=\angle FAE=\angle FKE=\angle DKM=\angle DAM.
This implies that ∠BAF=∠CAM\angle BAF=\angle CAM and ∠CAF=∠BAM\angle CAF=\angle BAM. It follows that
FBFC=sin⁡∠BAFsin⁡∠CAF=sin⁡∠CAMsin⁡∠BAM=AB⋅BMsin⁡∠AMBAC⋅CMsin⁡∠AMC=ABAC \frac{FB}{FC}=\frac{\sin \angle BAF}{\sin \angle CAF}=\frac{\sin \angle CAM}{\sin \angle BAM}=\frac{AB \cdot \frac{BM}{\sin \angle AMB}}{AC \cdot \frac{CM}{\sin \angle AMC}}=\frac{AB}{AC}
as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.