ABCDEFGH is a regular octagon. M is the midpoint of BF. How do you prove that triangles AMG and BDF are similar?

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1 Answer
Nov 12, 2017

see explanation.

Explanation:

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Given that #M# is the midpoint of #BF#, #M# is the center of the regular octagon.
central angle of the regular octagon : #angleAMB=angleBMC=....=angleGMH=angleHMA=360/8=45^@#,
#=> angleAMG=2xx45=90^@#
as #AM=GM, => DeltaAMG# is an isosceles right triangle.
Similarly, #DeltaBMD and DeltaFMD# are isosceles right triangles,
#=> angleMDB=angleMDF=(180-90)/2=45^@#
#=> angleBDF=45+45=90^@#,
as #BD=FD, => BDF# is also an isosceles right triangle.

Hence #DeltaAMG and DeltaBDF# are similar.