I'll just tell you how.  You can write up a two-column proof.

All sides of a regular polygon are congruent.  All interior angles of
a regular polynomial are congruent.

so AB ≅ AE, BC ≅ ED, ∠B ≅ ∠E, therefore ΔABC ≅ ΔAED by SAS, 
so AC ≅ AD by cpct.

Draw in diagonal CE.



Now ΔCDE ≅ ΔABC by SAS and AC ≅ CE

Draw in BE



Prove ΔCDE ≅ ΔABE the same way and then BE ≅ CE

Finally draw in BD



And now ΔBCD ≅ ΔABE and so BE ≅ BD.  Now write it up as a 
two-column proof.

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Are any two diagonals congruent in any regular polygon?

Take a look at a regular nonagon (nine sided regular polygon):



No way those two diagonals are congruent!  The only
regular polygons that have all diagonals congruent are
regular polygons with sides 4 and 5.  (A 4-sided
regular polygon is a square).

Edwin