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20 1 1 The answer to the last big math question in Mianyang senior high school entrance examination.
Solution: As shown in the figure, let AB=AC=2, then BC=2√2.

(1)∫d is ∴AD=CD= 1. The midpoint of communication.

In Rt△ABD, BD=√5 according to Pythagorean theorem.

And Rt△ABD∽Rt△ECD, so there are CE/CD=AB/BC, CE=AB*CD/BD=2/BD=2/√5.

∴BD/CE=BD/(2/BD)=BD^2/2=5/2.

(2)BD is the bisector of ∴ AD/CD = AB/BC = 2/(2 √ 2) = ∠ 2. ∠ B.

(What is used here is the nature of the angle bisector, which may not be mentioned in the current textbooks. )

(AD+CD)/CD=(2+√2)/2, that is, 2/CD=(2+√2)/2 and CD=2(2-√2).

AD = AC-CD = 2-2(2-√2)= 2(√2- 1)。 BD^2=AD^2+AB^2=8(2-√2).

∴ce=ab*cd/bd=4(2-√2)/bd. rt△Abd∽rt△ced

bd/ce=bd/{[4(2-√2)]/bd}=bd^2/[4(2-√2)]=[8(2-√2)]/[4(2-√2)]= 2。

(3) When D and A coincide, BD: CE = 1, taking the minimum value. When d approaches c, the ratio can be infinite, so the ratio is ≥ 1. So the ratio can be ≤4/3.