Problem: Angle B of triangle ABC is trisected by BD and BE which meet AC at D and E respectively. Then:
Answer Choices:
A. ECAD=DCAE
B. ECAD=BCAB
C. ECAD=BEBD
D. ECAD=(BE)(BC)(AB)(BD)
E. ECAD=(DC)(BE)(AE)(BD)
Solution:
Since BD bisects angle ABE, we have DEAD=BEAB;
since BE bisects angle DBC,
we have ECDE=BCBD.
Hence
ECAD=DE(BC/BD)DE(AB/BE)=(BE)(BC)(AB)(BD)
