Because you don't really know where point A should be relative to the segment BC, let me suggest the following: Take the perpendicular bisector of BC and at C construct half of angle A on one side of BC and the complement of half of angle A at C on the other side of BC. Suppose the new sides of those two angles meet the perpendicular bisector of BC at D and E respectively. Let O be the midpoint of segment DE. Then any point on the arc BEC of the circle with center O could be the point A, because an angle at a point on a circle measures half of the arc that it "subtends", so for any point A on that arc, angle BAC measures the same as angle BEC, which is the desired angle A. (To see this, recall that segment BC is perpendicular to DE -- say they meet at F -- and because we built angle FCE to be the complement of half of angle A, angle FEC is half of angle A, so angle BEC is equal to angle A.) The choice of point A on the arc BEC that gives the greatest area is the one that is the greatest perpendicular distance away from segment BC, and that is the point E, halfway along that arc. So, yes, the greatest area comes from an isosceles triangle: the lengths EB and EC are congruent.