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.
We can use following properties of Napoleon's Equilateral triangles.
ReplyDeleteAA'= 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.