∴FO=0.5CD=9,FO‖CD, that is, AO‖CD,∵PB=BO,∴PO=2/3PC times △AOP∽DCP.
Get AP = 2/3pd, ao = CD * 2/3 = 12, ∴ Bo = 12, ∫in rt△fbo, OF=9,OB= 12,∴BF? =63,
∫AF = AO-FO = 12-9 = 3, in Rt△ABF, AB=√(63+3? )=6√2=AD，
∫ AP = 2/3pd = 2/3ap+2/3 * 6 ∴ 2 ∴ AP =12: 2, let DE=x, then PE =18: 2+X, ∫ PC =
∴CE？ =36? -( 18√2+x)？ = 18? -x？ X=9√2/2, which means DE=9√2/2.
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