+0

# help

+1
47
1

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'. Jan 9, 2020

#1
+1

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

So

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   Jan 10, 2020