Traditional Culture Encyclopedia - Photography major - 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
∫∠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.
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