Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem.Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 998.
http://s25.postimg.org/8ixkdqthb/pro_998.pngConnect AE and BC Perform rotational transformation, center B, angle of Rot= 120Triangle CBD will become EBASo AE=DC and ∠ (DMA)=120We have FG=1/2AE=1/2DC=GH and ∠ (FGH)=60So triangle FGH is equilateral
Let z(P) be the complex number representing P. Let z(B)=0, z(D)=a, z(E)=b, denote ω=cis(120°). Thenz(A)=aω, z(C)=bω². z(F)=1/2 a(1+ω), z(H)=1/2 b(1+ω²), z(G)=1/2 (a+b)z(G) + ω z(F) + ω² z(H)= 1/2 (a+b) + 1/2 a(ω+ω²) + 1/2 b(ω²+ω)= 0Hence, FGH is equilateral.
sort of qualitative justification: see http://bleaug.free.fr/gogeometry/998.pnglet's consider I and J as centers of circumscribed triangles ABD and CBE. By construction, IDB and BEJ are equilateral therefore there exist a similarity transformation S that transform IDB into BEJ. Since T=(Id+S)/2 is also a similarity transformation (group + vector space over R), T(IDB)=FGH is equilateral.bleaug