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2條F.4數學問題 高手幫幫忙!!

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http://farm4.static.flickr.com/3020/3091791219_ecc72fd025.jpg?v=0 http://farm4.static.flickr.com/3155/3091790769_d85e576a6a.jpg?v=0 link入面有條數唔識做!!! help 詳細解答

最佳解答:

1. Since ∠ABE=∠DCE (∠s in the same segment) By the same reason, ∠BAE=∠CDE (∠s in the same segment) Also, ∠BEA=∠CED (vert. opp. ∠) Therefore ΔABE~ΔDCE (AAA) 2. Let ∠OAB=∠ACB=x, and join OB First, we have ∠AOB=2∠ACB=2x (∠ at centre twice ∠ at ⊙^ce) Also, since, OA=OB (radii of the same circle) thus ∠OBA=∠OAB=x (base ∠s, isos. Δ) Now consider ΔOAB, ∠OAB+∠OBA+∠AOB=180° (∠ sum of Δ) thus x+x+2x==4x=180°, hence ∠AOB=x=45°. 2008-12-09 18:19:20 補充： hotunghon咁抄都得啊？！搞錯啊！

其他解答:

• 問問關於英文tense 的問題@1@
• 咩係次按債券-
• tangent

 

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我知 <1. Since ∠ABE=∠DCE (∠s in the same segment) By the same reason, ∠BAE=∠CDE (∠s in the same segment) Also, ∠BEA=∠CED (vert. opp. ∠) Therefore ΔABE~ΔDCE (AAA) 2. Let ∠OAB=∠ACB=x, and join OB First, we have ∠AOB=2∠ACB=2x (∠ at centre twice ∠ at ⊙^ce) Also, since, OA=OB (radii of the same circle) thus ∠OBA=∠OAB=x (base ∠s, isos. Δ) Now consider ΔOAB, ∠OAB+∠OBA+∠AOB=180° (∠ sum of Δ) thus x+x+2x==4x=180°, hence ∠AOB=x=45°.>