Picture: http://postimg.org/image/kudu9b9tz/
In the figure, B is a point on the major arc AD such that BM is a perpendicular bisector of AD. AMGDF, BGE and BCF are straight lines. Prove that CFEG is a cyclic
quadrilateral.
Thank you!!
F.5 Maths circle(2)
2014-12-25 8:12 pm
回答 (2)
2014-12-26 6:33 am
✔ 最佳答案
圖片參考：https://s.yimg.com/rk/HA04628698/o/1406719366.jpg
Extend BM cut the circle at N , then BN is a diameter of the circle.
∠NEB = 90° (∠in semi-circle)
∠NMG = ∠AMB (vert. opp. ∠s) = 90° = ∠NEB ,
∴ NEGM is a cyclic quadrilateral.(opp. ∠s supp.)
Hence ∠MNE = ∠EGD (ext. ∠s, cyclic quad.)
On the other hand, ∠MNE = ∠ECF (ext. ∠s, cyclic quad. BNEC)
Therefore ∠EGD = ∠ECF.
Thus , CFEG is a cyclic (converse of ∠s in the same segment)
2014-12-26 05:08:03 補充：
Merry Christmas!
2014-12-26 9:49 am
Very tidy.
Thank you!
Merry Christmas!
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2014-12-26 05:31:49 補充：
★︵＿＿＿︵☆
╱ ╲
︴● ● ︴
︴≡ ﹏ ≡ ︴
╲＿＿＿＿＿╱
Thank you!
Merry Christmas!
╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\\╯
2014-12-26 05:31:49 補充：
★︵＿＿＿︵☆
╱ ╲
︴● ● ︴
︴≡ ﹏ ≡ ︴
╲＿＿＿＿＿╱
收錄日期: 2021-04-15 17:40:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141225000051KK00016