2002 AMC 10A Problems/Problem 20
Contents
Problem
Points 
and 
lie, in that order, on 
, dividing it into five segments, each of length 1. Point 
is not on line 
. Point 
lies on 
, and point 
lies on 
. The line segments 
and 
are parallel. Find 
.
Solution 1
First we can draw an image.
![[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, G, H, J; A = (0,0); B = (1,0); C = (2,0); D = (3,0); E = (4,0); F = (5,0); G = (-1.5,4); H = extension(D, G, C, C + G - A); J = extension(F, G, E, E + G - A); draw(A--F--G--cycle); draw(B--G); draw(C--G); draw(D--G); draw(E--G); draw(C--H); draw(E--J); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, S); label("$F$", F, SE); label("$G$", G, NW); label("$H$", H, W); label("$J$", J, NE); [/asy]](http://latex.artofproblemsolving.com/f/3/6/f36a37fb3fee1ec7e6c9eb3326ff0df1b01aedbe.png)
Since and 
are parallel, triangles 
and 
are similar. Hence, 
.
Since and 
are parallel, triangles 
and 
are similar. Hence, 
. Therefore, 
. The answer is 
.
Solution 2
As angle F is clearly congruent to itself, we get from AA similarity, 
; hence 
. Similarly, 
. Thus, 
.
See Also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 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 | ||
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