2009 AMC 10B Problem 9

Attempt Problem 9 of the 2009 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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9.

Segment BDBD and AEAE intersect at C,C, as shown, AB=BC=CD=CE,AB=BC=CD=CE, and A=52B.\angle A=\dfrac52\angle B. What is the degree measure of D?\angle D?

52.552.5

5555

57.557.5

6060

62.562.5

Answer: A
Concepts:angle chasingisosceles triangleangle sum
Difficulty rating: 1240
Solution:

Since ABC\triangle ABC is isosceles with AB=BC,AB=BC, we have A=C.\angle A=\angle C. With A=52B,\angle A=\dfrac52\angle B, the angle sum gives 52B+52B+B=180, \dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ, so B=30\angle B=30^\circ and ACB=75.\angle ACB=75^\circ.

By vertical angles DCE=75.\angle DCE=75^\circ. Since CD=CE,CD=CE, triangle CDECDE is isosceles, so 2D+75=180, 2\angle D+75^\circ=180^\circ, giving D=52.5.\angle D=52.5^\circ.

Thus, the correct answer is A.

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