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 diagram below to enlarge it.
(Actually copy from my last solution)Since AC₁ = AB₂.So AA₁ is the perpendicular bisector of C₁B₂,thus AC₁A₁B₂ is a kite, therefore A₁C₁=A₁B₂.Similarly, C₁A₁=C₁B₂.Hence, ΔB₂A₁C₁ is equilateral.
Another interesting fact: B₁ is actually the center of equilateral ΔB₂A₁C₁
Reference my proof in the previous problem, A1B2C1 is an isoceles triangle with included angle = 60Hence this triangle has to be equilateral Sumith PeirisMoratuwaSri Lanka