Track / Stage 6 / 145 of 400 #1145 of 1964

Problem 1145

National olympiad, first round

Geometry Difficulty 6.2 Prove it

Example 15 As shown in Figure 2.1.14, let circle $O$ be the excircle of $\triangle A B C$ opposite to side $B C$ , and let $D, E, F$ be the points of tangency of circle $O$ with $B C, C A$ , and $A B$ (or their extensions), respectively. If $O D$ intersects $E F$ at $K$ . Prove: $A K$ bisects $B C$ . (Refer to Example 13 in Chapter 11 of the first volume)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

From the figure, we know that $B, F, O, D$ are concyclic, so $\angle A B C=\angle F O D=\theta$ . Similarly, $\angle A C B=\angle D O E=\psi$ .

In $\triangle A B C$ , we have $\frac{A B}{A C}=\frac{\sin \phi}{\sin \theta}$ . Also, in $\triangle A F E$ and $\triangle O F E$ , using the Angle Bisector Theorem, we get
$\frac{F K}{K E}=\frac{A F \cdot \sin \alpha}{A E \cdot \sin \beta}=\frac{\sin \alpha}{\sin \beta}, \frac{F K}{K E}=\frac{O F \cdot \sin \theta}{O E \cdot \sin \psi}=\frac{\sin \theta}{\sin \psi}$

Thus,
$\frac{\sin \alpha}{\sin \beta}=\frac{\sin \theta}{\sin \psi}$

In $\triangle A B C$ , by the Angle Bisector Theorem,
$\frac{B M}{M C}=\frac{A B \cdot \sin \alpha}{A C \cdot \sin \beta}=\frac{\sin \psi}{\sin \theta} \cdot \frac{\sin \alpha}{\sin \beta}=1$

Therefore, $B M=M C$ , which means $A K$ bisects $B C$ .

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