Track / Stage 4 / 121 of 340 #381 of 2000

Problem 381

AMC 12 late, AIME early

Geometry Difficulty 4.7 Find the answer

$17 \cdot 99$ In triangle $ABC$ , point $F$ divides $AC$ in the ratio 1:2. Let $E$ be the intersection of side $BC$ and line $AG$ , where $G$ is the midpoint of $BF$ . Then the ratio in which $E$ divides side $BC$ is
(A) $1: 4$
(B) $1: 3$ .
(C) $2: 5$ .
(D) $4: 11$ .
(E) $3: 8$ .

Your answer

Official solution

[Solution] Draw $F H / / A E$ . Since $B G=G F$ , we have
$B E=E H \text {. }$

At the same time, from $2 A F=F C$ , we get $2 E H=H C$ .
$\therefore 3 B E=E H+H C=E C \text {. }$

Thus, $E$ divides $B C$ in the ratio $1: 3$ . Therefore, the answer is $(B)$ .

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