Problem:
Trapezoid ABCD has AD∥BC,BD=1,∠DBA=23∘, and ∠BDC=46∘. The ratio BC:AD is 9:5. What is CD ?
Answer Choices:
A. 97​
B. 54​
C. 1513​
D. 98​
E. 1514​
Solution:
Extend AB and DC to meet at E. Then
∠BED=180∘−∠EDB−∠DBE=180∘−134∘−23∘=23∘.​
Thus △BDE is isosceles with DE=BD. Because AD∥BC, it follows that the triangles BCD and ADE are similar. Therefore
59​=ADBC​=DECD+DE​=BDCD​+1=CD+1
so CD=54​.
\section*{OR}
Let E be the intersection of BC and the line through D parallel to AB. By construction BE=AD and ∠BDE=23∘; it follows that DE is the bisector of the angle BDC. By the Bisector Theorem we get
CD=BDCD​=BEEC​=BEBC−BE​=ADBC​−1=59​−1=54​​
The problems on this page are the property of the MAA's American Mathematics Competitions
