Track / Stage 6 / 45 of 400 #1045 of 2000

Problem 1045

National olympiad, first round

Geometry Difficulty 6.0 Prove it

1. In $\triangle ABC$ , let $AB > AC$ , and draw the tangent line $l$ to the circumcircle of $\triangle ABC$ at $A$ . Construct a circle with center $A$ and radius $AC$ , which intersects the line segment $AB$ at $D$ and the line $l$ at $E, F$ .
Prove: The lines $DE, DF$ pass through the incenter and one excenter of $\triangle ABC$ respectively.
Note: A circle that is tangent to one side of a triangle and the extensions of the other two sides is called an excircle of the triangle, and the center of the excircle is called an excenter.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Draw the angle bisector of $\angle B A C$ intersecting $D E$ at $I$ , and connect $I B$ , $I C, C D, C E$ . Since $A$ is the circumcenter of $\triangle D C E$ , then $\angle 1=\frac{1}{2} \angle B A C=\angle 2 \Rightarrow A, I, C, E$ are concyclic, thus $\angle C I E=\angle C A E=\angle A B C \Rightarrow I, D, B, C$ are concyclic, hence $\angle 3=\angle 6, \angle 4=\angle 5$ .
Also, $A D=A E \Rightarrow \angle 3=\angle 4 \Rightarrow \angle 5=\angle 6$ .
Therefore, $I$ is the incenter of $\triangle A B C$ . Extend $A I$ to intersect $C D$ at point $G$ , and the extension of $F D$ at point $I_{a}$ , and connect $B I_{a}$ . $A D=A E=A F \Rightarrow D E \perp D F, A C=A D \Rightarrow G A \perp G D$ , thus in the right triangle $\triangle I D I_{a}$ , $\angle 8=\angle 7=\angle 9=\angle I B D \Rightarrow I, D, B, I_{a}$ are concyclic, hence $\angle I B I_{a}=\angle I D I_{a}=90^{\circ} \Rightarrow I_{a}$ is an excenter of $\triangle A B C$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.