In triangle ABC, M is the midpoint of AB. Let D be the point on BC such that AD bisects angle BAC and let the perpendicular bisector of AB intersect AD at E. If AB = 44, AC = 30, and ME = 10 then find the area of triangle ACE.
Call the point, F, the foot of the perpendicular line from E to AC. Because triangles AFE and AME are congruent by AAS, FE = 10, and FA = 22. This gives an area of 110. Because AC = 30, FC = 30 - 22 = 8. The area of triangle EFC is therefore 40.
Adding the two triangle areas up gives us 150 units squared.