Problem:
In the obtuse triangle ABC,AM=MB,MDβ₯BC,ECβ₯BC. If the area of β³ABC is 24, then the area of β³BED is
Answer Choices:
A. 9
B. 12
C. 15
D. 18
E. not uniquely determined
Solution:
Draw in MC as in the adjoining figure. Then β³DMC and β³DME have the same area, as they have the same base MD and equal altitudes on that base. Thus
Area β³BMC= Area β³BED
Moreover, as shown below,
Area β³BMC=21β Area β³BAC
Thus
Area β³BED=21βArea β³BAC=12.
As for the claim in the second display, since M is the midpoint of BA, altitude MD of β³BMC is one half altitude AF of β³BAC. Since the triangles have the same base BC, the claim follows.
Alternately, one proves the claim by noting that
Area β³BMC=21β(BM)(BC)sinBβ=21β[21β(BA)(BC)sinB]=21β Area β³BACβ