The bisector of the inner and outer angles. So < ECF=90 degrees. To be a rectangle, the diagonal lines should be equally divided.

-& gt； AO=OC。

That is, o is the midpoint of AC. AECF is rectangular.

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Similarly, fo = oc-> EO=FO and ao = oc- > diagonal bisection. AECF is a parallelogram. < ECF = 90-> again; AECF is rectangular.

Must be equal. The above proof proves that/

EF//BD-& gt； & ltBCE =< The CEO is

Similarly, fo = oc-> EO=FO.