The moving point problem of mathematics and geometry in the second day of junior high school

1) quadrilateral DEBF is a parallelogram. ∫e and f moving beams are equal, so OE = ao-AE = oc-cf = of od = ob ∠ EOD = ∠ FOB ∴ de = BF ∠ deo = ∠ BFO. Prove 2)BD= 12cm, BO=6cm, when OF=OC-FC=OE=OA-EA=OB=6cm ∴ when T=2, there is FC=EA=2cm, if OF=OE=6cm, then quadrilateral D, E and B.

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