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Shijingshan Grade Three Mathematical Model II
(1) Proof: connect PO and PC, as shown in the figure.

∵BC is the diameter of⊙ O,

∴∠BPC=90,

∴∠APC=90,

Q is the midpoint of AC,

∴PQ=CQ,

∴∠CPQ=∠PCQ,

OP = OC

∴∠OPC=∠OCP,

∴∠OPC+∠CPQ=∠OCP+∠PCQ=∠BCA=90,

∴OP⊥PQ,

∴ The straight line PQ is tangent to⊙ O; 、

(2) Solution: Connect CE, as shown in the figure.

∫EP is the diameter,

∴∠ ECP = 90, that is ∠ ECO+∠ OCP = 90,

∠∠ Ecology +∠ ECF = 90,

∴∠ECF=∠OCP=∠OPC,

And < f = < f.

∴△FEC∽△FCP,

∴EFCF=CFPF=ECPC,

In Rt△EPC, tan∠OPC=ECPC= 12.

∴EFCF=CFPF= 12,

∴CF=2EF,PF=2CF,

∴PF=4EF,

∴PE=3EF,

That is, 3EF=2×5,

∴EF=253.