Problem: In â–³ABC with right angle at C, altitude CH and median CM trisect the right angle. If the area of â–³CHM is K, then the area of â–³ABC is
Answer Choices:
A. 6K
B. 43​K
C. 33​K
D. 3K
E. 4K
Solution:
Tringles ABC,AMC,MBC,MBC and HBC all haw the same altitude HC so that their areas (denoted below by parentheses) are proportional to their bases. We have
(AMC)=(MBC) because AM=MB. Also (MBC)=(MHC)+(HBC)=K+K=2K because
△MHC and △HBC are congruent 30∘−60∘ right triangles.
