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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
http://s16.postimg.org/u34puqelh/pro_1160.pngDenote S(XYZ)= area of XYZDraw altitudes AR and CL and lines per sketchObserve that triangle B1HC congrurent to CLA ( case ASA) => B1H=CLWe have S(B1BC)= ½ .(BC1 x B1K).... (1)But B1K= B1H+HK= CL+ BLReplace this in (1) we have S(B1BC)=1/2. BC1 x(CL+BL)= S(ABC)+1/2. S(BLMC1)Simillarly S(B2BA2)= S(ABC)+ ½. S(BRSA2)Note that quadrilateral ACRL is cyclic => BL. BA=BR.BC => S(BLMC1)= S(BRSA2)Replace it in above expressions we get S(B1BC)= S(B2BA2)
I doubt I would ever come up with such a brilliant proof like Peter's. I proved this with a somewhat dumber approach using algebra.Area(BB1C1) = BB1*B1C1*sin(∠BB1C1)/2Area(BB2A2) = BB2*A2B2*sin(∠BB2A2)/2My approach involves these two steps:(1) BB1*B1C1 == BB2*A2B2(2) ∠BB1C1 == ∠BB2A2They are not trivial but not too hard to reach either.