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Difference between revisions of "2009 AMC 10B Problems/Problem 9"
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Latest revision as of 16:39, 28 June 2021

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 $\angle D$?

[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), Ep=dir(35), D=dir(-35), B=dir(145); pair A=intersectionpoints(Circle(B,1),C--(-1*Ep))[0]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(A--Ep--D--B--cycle); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,N); label("$E$",Ep,E); label("$D$",D,E); [/asy]

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

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 $\frac{180^\circ - 75^\circ}2 = \frac{105^\circ}2 = \boxed{52.5}$.

Video Solution

https://youtu.be/hsP804ZSocg

~savannahsolver

See Also

2009 AMC 10B (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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