In triangle ABC, M is the midpoint of AB. Let D be the point on BC such that AD bisects BAC, and let the perpendicular bisector of AB intersect AD at E. If AB=44, AC=30, and ME=\(20\), then find the area of triangle ACE
First I drew out all the instructions. AME is a right triangle because it intersects AD with a perpendicular bisector. Then I drew a line down from E to AC that was perpendicular to AC. That length is the height of triangle ACE, and AC is the base, which is 30. So we have to find that height. Let us call the intersection of the orthogonal line segment from E to AC, P.
ME is 20, and AME is a right triangle. AD is a perpendicular bisector, so angle CAE is congruent to angle MAE. Angle AME is congruent to angle APE. AE is congruent to AE using reflexive property, so triangle APE is congruent to triangle AME. Thus PE, the height of triangle ACE with base AC, is 20 (because ME is 20).
Now since we know the base of triangle ACE is 30, and the height is 20, then we can use (Base x Height)/2 to get the area.
Thus, the area of triangle ACE is 300 units squared.