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)​
