-& gt; AO=OC。
That is, o is the midpoint of AC. AECF is rectangular.
EF//BD-& gt; & ltBCE =< The CEO is
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.