Difference between revisions of "2002 AMC 10A Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>. | + | As <math>\overline{JE}</math> is parallel to <math>\overline{AG}</math>, angles FHD and FGA are congruent. Also, angle F is clearly congruent to itself. From SSS similarity, <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 09:44, 27 December 2008
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
As 
is parallel to , angles FHD and FGA are congruent. Also, angle F is clearly congruent to itself. From SSS 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 | ||
