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

enter image source here

1 Answer
Nov 12, 2017

see explanation.

Explanation:

enter image source here
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.