luong@tonto.DEC (Van Luong Nguyen UHO DTN 264-6560) (07/17/84)
>Subject: Geometry V >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 There must be a mistyping in the above problem: As stated, point E is indeterminate... The following solution ASSUMES that E is a point on MA such that EC is prpndclr to BC. Then, denoting by (XYZ) the area of a triangle XYZ: (MCB) = 0.5 BC.MD and (ECB) = 0.5 BC.EC therefore: (MCB)/(ECB) = MD/EC = DB/CB (since MD and EC are parallel) (1) Draw perpendiculars CC' and DD' to EB. Now: (DEB) = 0.5 EB.DD' and (ECB) = 0.5 EB.CC' Therefore: (DEB)/(ECB) = DD'/CC' = DB/CB (since DD' and CC' are parallel) (2) Combining (1) and (2) shows that: (DEB) = (MCB) (3) But (MCB) = 0.5 CC'.MB (MCB) = 0.5 CC'.(AB/2) = (0.5 CC'.AB)/2 = (ABC)/2 = 12 (4) Finally, combining (3) and (4): (DEB) = 12 ========== Van Luong Nguyen, Digital Equipment Corporation.