Track / Stage 5 / 136 of 400 #736 of 2000

Problem 736

AIME late

Geometry Difficulty 5.4 Prove it

Five. (16 points) As shown in Figure 2, given that the circumcenter of $\triangle ABC$ is $O$ , draw any circle passing through points $B$ and $C$ , intersecting the extensions of $AB$ and $AC$ at points $E$ and $F$ respectively. Prove: $AO \perp EF$ .

保留源文本的换行和格式，直接输出翻译结果如下：

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

As shown in Figure 6, construct the circumcircle $\odot O$ of $\triangle A B C$ , and extend $A O$ to intersect $\odot O$ and $E F$ at points $D$ and $G$ respectively. Connect $C D$ . It is easy to see that
$\begin{array}{l} \angle A E G=\angle A C B, \\ \angle B A D=\angle B C D, \angle A C D=90^{\circ} . \end{array}$

Thus, $\angle A E G+\angle E A G=\angle A C B+\angle B C D$
$=\angle A C D=90^{\circ} \text {. }$

Therefore, $\angle A G E=90^{\circ}$ .
Hence, $A G \perp E F$ , which means $A O \perp E F$ .

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