# 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 $HC/JE$.

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

$[asy] pair A,B,C,D,EE,F,G,H,J; A = (0,0); B = (0.2,0); C = 2*B; D = 3*B; EE = 4*B; F = 5*B; G = (-0.2,0.8); H = intersectionpoint(G--D,C -- (C + G)); J = intersectionpoint(G--F,EE--(EE+G)); draw(G--F--A--G--B); draw(H--C--G--D); draw(J--EE--G); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",EE,S); label("$F$",F,SE); label("$J$",J,NE); label("$G$",G,N); label(scale(0.9)*"$H$",H,NE,UnFill(0.1mm)); [/asy]$

## Solution

Solution $\text{#1}: Since$ (Error compiling LaTeX. ! You can't use `macro parameter character #' in restricted horizontal mode.)AG$and$CH$are parallel, triangles$GAD$and$HCD$are similar. Hence,$CH/AG = CD/AD = 1/3$. Since$ (Error compiling LaTeX. ! Missing $inserted.)AG$and$JE$are parallel, triangles$GAF$and$JEF$are similar. Hence,$EJ/AG = EF/AF = 1/5$. Therefore,$CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}$. The answer is (D).

Solution \text{#2}: As$(Error compiling LaTeX. ! Missing$ inserted.)\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)}}\$.