# Difference between revisions of "2002 AMC 10A Problems/Problem 20"

## Problem

Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$.

$\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2$

## Solution 1

Since $\overline{AG}$ and $\overline{CH}$ are parallel, from AA similarity, triangles $GAD$ and $HCD$ are similar. Hence, $\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}$.

Since $\overline{AG}$ and $\overline{JE}$ are parallel, from AA similarity, triangles $GAF$ and $JEF$ are similar. Hence, $\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}$. Therefore, $\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}$. The answer is $\boxed{(D)}$.

## Solution 2

As $\overline{JE}$ is parallel to $\overline{AG}$, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, $\triangle AGF \sim \triangle EJF$; hence $\frac {AG}{JE} =5$. Similarly, $\frac {AG}{HC} = 3$. Thus, $\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$.

## See Also

 2002 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS