2003 AMC 10A Problems/Problem 22
Problem
In rectangle , we have
,
,
is on
with
,
is on
with
, line
intersects line
at
, and
is on line
with
. Find the length of
.
Solution
Solution 1
Since is a rectangle,
.
Since is a rectangle and
,
.
Since is a rectangle,
.
So, is a transversal, and
.
This is sufficient to prove that and
.
Using ratios:
Since can't have 2 different lengths, both expressions for
must be equal.
Solution 2
Since is a rectangle,
,
, and
. From the Pythagorean Theorem,
.
Lemma
Statement:
Proof: , obviously.
$\begin{eqnarray} \angle HCE=180^{\circ}-\angle CHG\\ \angle DCE=\angle CHG-90^{\circ}\\ \angle CEED=180-\angle CHG\\ \angle GEA=\angle GCH \end{eqnarray}$ (Error compiling LaTeX. Unknown error_msg)
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
Let .
Also, , therefore
We can multiply both sides by to get that
is twice of 10, or
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AMC 10 Problems and Solutions |