Question

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

Answer 1

see explanation.

Explanation:

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.