2011 AMC 10B Problem 17

Attempt Problem 17 of the 2011 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 2011 AMC 10B solutions, or check the answer key.

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17.

In the given circle, the diameter EB\overline{EB} is parallel to DC,\overline{DC}, and AB\overline{AB} is parallel to ED.\overline{ED}. The angles AEBAEB and ABEABE are in the ratio 4:5.4 : 5. What is the degree measure of angle BCD?BCD?

120120

125125

130130

135135

140140

Answer: C
Concepts:inscribed angleparallel linesangle chasing
Difficulty rating: 1670
Solution:

Since EBEB is a diameter, EAB=90\angle EAB=90^\circ. The ratio AEB:ABE=4:5\angle AEB:\angle ABE=4:5 then gives AEB=40\angle AEB=40^\circ and ABE=50\angle ABE=50^\circ.

Because ABEDAB\parallel ED, DEB=50\angle DEB=50^\circ. Since EBDCEB\parallel DC, quadrilateral EBCDEBCD is an isosceles trapezoid, so BCD=CDE\angle BCD=\angle CDE.

Angles DEBDEB and CDECDE are supplementary, so CDE=18050=130\angle CDE=180^\circ-50^\circ=130^\circ. Hence BCD=130\angle BCD=130^\circ.

Thus, C is the correct answer.

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