CN-SMO 2006 Problem 2

In $\triangle ABC$ , $\angle ABC=90^{\circ}$ . Points $D,G$ lie on side $AC$ . Points $E, F$ lie on segment $BD$ , such that $AE \perp BD$ and $GF \perp BD$ . Show that if $BE=EF$ , then $\angle ABG=\angle DFC$ .