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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1205.
We can use following properties of Napoleon's Equilateral triangles. AA'= BB'= CC' And Triangle A1B1C1 is equilateral triangle Consider D as midpoint of side AC, In Triangle CDC', C1D=C'D/3 And GD=CD/3.Hence line joining C1G is parallel to CC' and also C1G=CC'/3 Similarly B1G=BB'/3 And A1G=AA'/3 We get A1G=B1G=C1G, hence G is circumcentre as well as centroid of Triangle A1B1C1 (its equilateral ) Similarly we can prove for Triangle A2B2C2.