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Geometry question

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How to do this question? 圖片參考:https://s.yimg.com/rk/HA04864973/o/2090329704.jpg 更新: Please go to the site below to solve Geometry question 4 if you can: https://hk.knowledge.yahoo.com/question/question?qid=7015030400088

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Int.∠ of a regular octagon = 180° - (360°/8) = 135° In ΔBCD : CB = CD (regular octagon) ∠CDB = ∠CBD= x (base ∠s of isos. Δ) ∠C + ∠CDB + ∠CBD= 180° (∠sum of Δ) 135° + x + x =180° x = 22.5° ΔABH ? ΔCBD (SAS) ∠ABH = x = 22.5° (corr. ∠s) In quad. ABGH : ∠A = ∠AHG = 135° and AB = HG Hence, ABGH is an isos. trapezium, i.e. AH // BG ∠A + ∠ABG = 180° (int. ∠s,AH // BG) 135° + ∠ABG = 180° ∠ABG = 45° z = ∠ABG- ∠ABH= 45° - 22.5° = 22.5° By the symmetric property of regular octagon, ΔBGF ?BGF ∠GBF = ∠EBF = y (corr. ∠s) By the symmetric property of regular octagon, ∠ABF= (1/2)∠ABC ∠ABH + ∠HBG + ∠GBF = (1/2) * 135° x + y + z = (1/2) * 135° x + y + z = 67.5° ...... The answer : B

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