Problem:
In the adjoining diagram, BO bisects β CBA,CO bisects β ACB, and MN is parallel to BC. If AB=12,BC=24, and AC=18, then the perimeter of β³AMN is
Answer Choices:
A. 30
B. 33
C. 36
D. 39
E. 42
Solution:
Since MN is parallel to BC,
β MOB=β CBO=β OBM,
β CON=β OCB=β NCO.
Therefore, MB=MO and ON=NC, and AM+MO+ON+AN
=(AM+MB)+(AN+NC)
=AB+AC=12+18=30.