gupta@asgb.UUCP (07/26/84)
>Once again, I have another problem, although could be one of the last. Here >goes: In the obtuse triangle ABC (angle C is obtuse), point M is on AB such >that AM = MB, point D is on BC such that MD is prpndclr to BC, and point E >is a point on ME such that EC is prpndclr to BC. Given the area of triangle >ABC is 24, then find area of triangle BED. Good luck! >Billy Pizer Correction> >E is a point on MA such that EC is perpendicular to BC. Solution> As M is the mid-point of AB, area(MCB) = area(MCA) = 1/2 area(ABC) = 12 ... (1) Also, area(DMC) = area(DME) ... (2) because they are on the same base (DM) and have the same height (DM is parallel to CE). Now, area(MCB) = area(MDB) + area(DMC) = area(MDB) + area(DME) ... (from 2) = area(BED) . . . area(BED) = 12 ... (from 1) Yogesh Gupta sdcrdcf --- bmcg!asgb!gupta sdcsvax ---/