Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 771.
http://img266.imageshack.us/img266/9572/problem771.pngAssign areas per attached sketchWe have Area(EBC)=S2+S3+P1= 1/2. BC*h1Area(EAD)=S4+S5+P2= ½. AD * h2So S2+S3+S4+S5+P1+P2=1/2. BC*(h1+h2)= ½.BC*h = ½ Area(ABCD)Area(ABG)=1/2*h*BG=Area(AGF) => S1=S6+P1+P2Combine above 2 expressions we get S1+S2+S3+S4+S5-S6= ½.Area(ABCD)
Let the area of "white" triangle with a vertex at F and at G be a and b respectively. First, S(BCE) + S(ADE) = S/2, (S2 + b + S3) + (S4 + a + S5) = S/2(S2 + S3 + S4 + S5) + (a + b) = S/2.....(1)Now consider trapezium ABGF, let AG and BF intersect at H. Then S(ABH) = S(FGH), S1 = S6 + (a + b)a + b = S1 - S6.....(2)Substitute (2) into (1), S1 + S2 + S3 + S4 + S5 - S6 = S/2