Competition problem — Geometry — Maths Olympiad Prep

Library / /402 of 520

Geometry Difficulty 5.7 AIME, harder Prove it

一、(40 points) As shown in Figure 1, given that the circle $\odot O$ with diameter $BC$ intersects the sides $AC$ and $AB$ of $\triangle ABC$ at points $D$ and $E$ respectively, $BD$ and $CE$ intersect at point $F$ , and $G$ is a point on segment $DE$ . $AI \perp FG$ , intersecting $FG$ at point $H$ , and intersecting $OD$ at point $I$ . Prove: $B$ , $G$ , and $I$ are collinear.

Solution

Connect $B I$ , $E I$ , and $C I$ . Then,
$B$ , $G$ , and $I$ are collinear $\Leftrightarrow \frac{S_{\triangle B E I}}{S_{\triangle B D I}}=\frac{E G}{G D}$ .
Since $O$ is the midpoint of $B C$ , we have
$S_{\triangle B D I}=S_{\triangle C D I}$ .
And since $B C$ is the diameter, hence
$\angle C E A=\angle B D A=90^{\circ}.$

Combining $A I \perp F G$ , we have $A$ , $E$ , $F$ , $D$ , and $H$ are concyclic. Thus,
$\begin{array}{l} \angle B A I+\angle E F G=180^{\circ}, \\ \angle C A I=\angle D F G. \\ \text{Then } \frac{S_{\triangle B E I}}{S_{\triangle B D I}}=\frac{S_{\triangle B E I}}{S_{\triangle C D I}}=\frac{B E \cdot A I \sin \angle B A I}{C D \cdot A I \sin \angle C A I} \\ =\frac{\sin \angle B A I \cdot \sin \angle B D E}{\sin \angle C A I \cdot \sin \angle C E D} \\ =\frac{\sin \angle E F G \cdot \sin \angle F D E}{\sin \angle D F G \cdot \sin \angle F E D}=\frac{E G}{G D}. \end{array}$

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