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2009 AMC 10B Problems/Problem 9

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Problem

Segment $BD$ and $AE$ intersect at $C$ , as shown, $AB=BC=CD=CE$ , and $\angle A = \frac 52 \angle B$ . What is the degree measure of $\angle D$ ?

$\text{(A) } 52.5\qquad\text{(B) } 55\qquad\text{(C) } 57.7\qquad\text{(D) } 60\qquad\text{(E) } 62.5$

Solution

$\triangle ABC$ is isosceles, hence $\angle ACB = \angle CAB$ .

The sum of internal angles of $\triangle ABC$ can now be expressed as $\angle B + \frac 52 \angle B + \frac 52 \angle B = 6\angle B$ , hence $\angle B = 30^\circ$ , and each of the other two angles is $75^\circ$ .

Now we know that $\angle DCE = \angle ACB = 75^\circ$ .

Finally, $\triangle CDE$ is isosceles, hence each of the two remaining angles ( $\angle D$ and $\angle E$ ) is equal to $\frac{180^\circ - 75^\circ}2 = \frac{105^\circ}2 = \boxed{52.5}$ .

See Also

2009 AMC 10B (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
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

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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

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