Difference between revisions of "2009 AMC 12B Problems/Problem 16"
m (→Solution 2: Fixed spelling error in "construction".) |
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=== Solution 2 === | === Solution 2 === | ||
− | Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line through <math>D</math> parallel to <math>\overline {AB}.</math> By | + | Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line through <math>D</math> parallel to <math>\overline {AB}.</math> By construction <math>BE = AD</math> and <math>\angle BDE = 23^{\circ}</math>; it follows that <math>DE</math> is the bisector of the angle <math>BDC</math>. So by the [[Angle Bisector Theorem]] we get |
<cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath> | <cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath> | ||
The answer is <math>\mathrm{(B)}</math>. | The answer is <math>\mathrm{(B)}</math>. |
Revision as of 14:16, 31 December 2018
Problem
Trapezoid has
,
,
, and
. The ratio
is
. What is
?
Solution
Solution 1
Extend and
to meet at
. Then
Thus is isosceles with
. Because
, it follows that the triangles
and
are similar. Therefore
so
Solution 2
Let be the intersection of
and the line through
parallel to
By construction
and
; it follows that
is the bisector of the angle
. So by the Angle Bisector Theorem we get
The answer is
.
See also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.