Problem:
n the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4:5. What is the degree measure of angle BCD?
Answer Choices:
A. 120
B. 125
C. 130
D. 135
E. 140
Solution:
Angle EAB is 90∘ because it subtends a diameter. Therefore angles BEA and ABE are 40∘ and 50∘, respectively. Angle DEB is 50∘ because AB is parallel to ED. Also, ∠DEB is supplementary to ∠CDE, so ∠CDE= 130∘. Because EB and DC are parallel chords, ED=BC and EBCD is an isosceles trapezoid. Thus ∠BCD=∠CDE=(C)130∘​.
OR
Let O be the center of the circle. Establish, as in the first solution, that ∠EAB= 90∘,∠BEA=40∘,∠ABE=50∘, and ∠DEB=50∘. Thus AD is a diameter and ∠AOE=100∘. By the Inscribed Angle Theorem
∠BCD=21​(∠BOA+∠AOE+∠EOD)=21​(80∘+100∘+80∘)=(C)130∘​.
The problems on this page are the property of the MAA's American Mathematics Competitions
.jpg)