Maths Question

Scarbo

2010-10-01 10:55 pm

圖片參考：http://imgcld.yimg.com/8/n/HA00135381/o/701010010082113873374760.jpg

1. In the figure, chords AB and ED of the circle are produced to meet at C. It is given that EA=EB.
a) Prove that angle BCE =angle EAD.
b) Is it possible that triangle BCE similar to EAD? Explain your answer.
ANS: NO

Thanks!

回答 (1)

用戶2053

2010-10-02 12:16 am

✔ 最佳答案

a)Joint BD ,ㄥEAB = ㄥEBA (base ∠s, isos.Δ) ㄥEBD = ㄥEAD (∠s in the same segment)So ㄥDBA = ㄥEBA + ㄥEBD ㄥDBA = ㄥEAB + ㄥEAD ....(1)ㄥBDC = ㄥEAB (ext. ∠s, cyclic quad.) ;ㄥBCE + ㄥBDC = ㄥDBA (ext. ∠ of Δ) i.e. ㄥBCE + ㄥEAB = ㄥDBA (Since ㄥBDC = ㄥEAB ) ....(2)(1) + (2) :ㄥDBA + ㄥBCE + ㄥEAB = ㄥEAB + ㄥEAD + ㄥDBA ㄥBCE = ㄥEADb)No.Since ㄥBCE = ㄥEAD is proved , and ㄥBEC < ㄥAED ,so ㄥBEC must be equal to ㄥEDA if △ BCE similar to △EAD.But ㄥEDA = ㄥEBA (∠s in the same segment) = ㄥBEC + ㄥBCE (ext. ∠ of Δ) which must be > ㄥBEC.So it is impossible.

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