Question

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 $\frac{HC}{JE}.$

$\dfrac{5}{4}$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

Answer: D

Concepts:similarity parallel lines

Difficulty rating: 1460

Small Hint:

$\triangle DHC\sim\triangle DGA$ gives $HC$ in terms of $AG$

Big Hint:

$\triangle FJE\sim\triangle FGA$ gives $JE$ in terms of $AG$

Solution:

Since $HC\parallel AG,$ $\triangle DHC\sim\triangle DGA,$ so $\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3},$ giving $HC=\dfrac{AG}{3}.$

Since $JE\parallel AG,$ $\triangle FJE\sim\triangle FGA,$ so $\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5},$ giving $JE=\dfrac{AG}{5}.$

Therefore $\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}.$

Thus, the correct answer is D.

2002 AMC 10A Problem 20

Answer: D

Difficulty rating: 1460

Small Hint:

Big Hint:

Solution:

Problem 20 in Other Years