As shown in the figure, in rtδABC, ∠ ABC = 90, AB=BC, ⊙O intersects OC at D with the diameter of AB, the extension line of D, AD intersects BC at E, and the tangent DF of ⊙O intersection D intersects

As shown in the figure, in rtδABC, ∠ ABC = 90, AB=BC, ⊙O intersects OC at D with the diameter of AB, the extension line of D, AD intersects BC at E, and the tangent DF of ⊙O intersection D intersects BC at F. (1) ∵ df is the tangent of ⊙O, ∴OD⊥DF,

∫∠ABC = 90，OF=OF，OD=OB，

∴rtδofd≌rtδofb(hl),∴∠fod=∠fob，

∵OA=OD,∴∠OAD=∠ODA，

And ∠BOD=∠OAD+∠ODA=2∠OAD,

∴∠FOB=∠OAD,∴OF∥AE。

(2) Let the radius of ⊙O be r, then BC=AB=2R, OC = √ (ob 2+BC 2) = √ 5r,

∴CG=CD=√5R-R=(√5- 1)R，

∴G ∴CG/CB=(√5- 1)/2 is a prime location in BC.

(3) connecting BD and H, where ∫ab is the diameter, ∴BD⊥AE,

∴δcdf∽δcbo (right angle, common angle),

∴df/cd=ob/bc= 1/2,∴df= 1/2cd= 1/2(√5- 1)r，

∵BC is also the tangent of⊙ O and∴ DF = BF, so DF is the center line of RT δ BDE.

∴BE=2DF=(√5- 1)R，

According to the projective theorem:

BE^2=DE*AE,AB^2=AE*AE，

∴de/ae=be^2/ab^2=(√5- 1)^2r^2/(4r^2)=(6-2√5)/4=(3-√5)/2

Projective theorem can be obtained from the similarity of right triangle. Because there are too many steps, it is replaced by projective theorem.