It is proved that connecting BE, crossing E is parallel to BF, crossing F is parallel to BE, two straight lines cross G point, connect AG, and then extend BE to cross AG at H point. rule

BF=EG，∠AEG=∠ECF

At △AEG and △ECF, AE/EC=EG/CF, ∠AEG=∠ECF,

So: two triangles are similar.

So: ∠EAG=∠CEF

So: AG‖DF

So: quadrilateral EFGH and quadrilateral BFGE are parallelograms.

So: BE=FG=EH

And ag-de.

So: AD=BD