ABCDE is a regular pentagon . A'B'C'D'E' is the pentagon inscribed in ABCDE's diagonals. Find similarity ratio of ABCDE to A'B'C'D'E'.
Note that angle BAC = 108°
And diagonals AD and AE trisect this angle
And note that angle BAD = BDA = 72°
So in triangle BAD' , AB = BD'
Let AB = one side of the pentagon = S = BD'
And let BE' = 1
And note that AB > BE'
Also note that AE' = BE'
And since angle BAE' = 36°, then angle ABE' also =36“
And, by symmetry, angles ABC and ACB also = 36°
So....by Angle-Angle congruency triangle BE'A is similar to triangle BAC
And note that BC = BD' + D'C = BD' + BE' = S + 1
BE' / AB = BA / BC or, put another way
1/ S = S / [ S + 1] cross-multiply
S^2 = S + 1 rearrange as
S^2 - S = 1 complete the square on S
S^2 - S + 1/4 = 1 + 1/4
(S - 1/2)^2 = 5/4 take the positive root
S - 1/2 = √5/ 2 add 1/2 to both sides
S = [ 1 + √5 ] / 2 [ this result is known as Phi ≈ 1.618 ]
So BD' = Phi
And BE' = 1
And E'D' = BD' - BE' ......so E'D' = Phi - 1
And we have the identity that
Phi - 1 = phi = E'D' where phi = the reciprocal of Phi ≈ .618
And E'D' is a side of the pentagon A'B'C'D'E' which is similar to the the larger pentagon ABCDE
So....the similarity ratio = AB / E'D' = Phi / phi = Phi / ( 1 / Phi) = Phi^2