Pass through point B as BM⊥BF and pass through CF at point M.
⑵∵AF⊥CF,AB⊥BC,
Easy to get: ∠FAB=∠BCF.
And ∠∠FBM =∠ABC = 90, we can get: ∠ABF=∠BCM, AB=AC.
∴△ABF≌△BCM(ASA), so BF=BM, △BFM is an isosceles right triangle.
∴∠BFC=∠BMF=45。
∫BF∨CD
∴∠DCF=∠BFC=45。
From isosceles right angle △ADE and isosceles right angle △ABC, we can get △ ADC △ ace (SAS),
∴∠ACD=∠ACF=22.5,
From the isosceles right angle △ABC: ∠ BCF = 45,
∴∠BCF=∠ACF=∠ACD=22.5。
∴CE sharing∠ ∠ACB
(1) from (1): ∠ BMF = 45,
∴∠CBM=45 -∠BCF=22.5。
∠CBM=∠BCF,
So BM=CM=BF,
Easy to get: EBM = BEM = 90-22.5 = 67.5,
∴EM=BM
Therefore, CE=2BM=2BF.
If you are an eighth-grade student, give the best answer!
What courses are there for education majors?
1, principles of pedagogy
Wang Daojun Wang Hanlan: Pedagogy, People's Ed