Given that each side of a convex pentagon is parallel to one of its diagonals. What may be the ratio of the length of a side to that of the parallel diagonal?

Here is my proof (you can apply this proof with any side of any pentagon):

Call the pentagon ABCDE. Call the length of CD, k. We have to find the length of the parallel diagonal of CD - which is BE.

Draw the pentagon. AC meet BE at M. Call the length of BM, x. BD meet AC at N, meet CE at H.

Because BM = x we have that BE = x + k
Because BE // CD so:
BM/CD = BN/ND = HD/BH = CD/BE
--> BM.BE = CD² --> x x (x+k) = k²
which means
BE = CD² /BM

NOTE: when k is known, we can find the value of x from the formula x x (x+k) = k² by solving the equation:
k² + kx - x² = 0

Ngoc,

There is also an approach by linear transformations, which avoids you having to do any real work at all!

Given the trapezoid ABCD (where AD // BC), one can apply a shear parallel to AD and BC which makes it into an isosceles trapezium. Assume AD is longer than BC.

Define K = intersection of line through A parallel to BD and line through D parallel to AC, and define K' = intersection of line thru B par to CD and line thru C par to AB.

Evidently as the angle ABC increases, K will move away from BC and K' will move towards it. Hence there is a unique point at which K = K' (and it is not difficult to see that at this unique point ABCDE is a regular pentagon stretched parallel to BC).

David

$x\times (x+k)=k^2$, let $x$ be the variable. So:

$x=k\times (\sqrt{5}-1)/2$ ( $x=CD$)

$BE=k+x=k\times (\sqrt{5}+1)/2$,

$CD/BE=k/(k\times (\sqrt{5}+1)/2$

$CD/BE=1/(\sqrt{5}+1)/2$

I think the ratio is the same in all case: any side/diagonal, in every possible convex pentagon.
It was a deep thought to use the ratios (BM/CD = BN/ND = HD/BH = CD/BE)to get the solution. I didn't see this.

I thank everybody for the help!

Peter