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
2026-03-25 13:57:02.1774447022
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
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Because by similarity we obtain: $$\frac{FY}{YB}=\frac{EF}{BC}=\frac{EF}{AD}=\frac{FX}{XA}.$$