# ABCD and AECF are two parallelograms and side EF is parallel to AD . suppose AF and DE met at X and BF AND CE AT Y . prove that XY is parallel to AB

I tried proving it by showing angles exy and eyx equal to edc and ecd respectively but I got no where . Is there any other approach I should consider

• Please edit title and body of the question to make it more readable. You could at least use GeoGebra to make a decent diagram. – Aretino Mar 15 at 8:17
• HINT: use intercept theorem. – Aretino Mar 15 at 8:21
• HINT: ...and a ruler. – TonyK Mar 15 at 9:59

Because by similarity we obtain: $$\frac{FY}{YB}=\frac{EF}{BC}=\frac{EF}{AD}=\frac{FX}{XA}.$$