我想要中一的幾何題,40分!

圖片參考：http://i169.photobucket.com/albums/u209/lautszki/F.jpg
Find the following angles in only one step:
i) ∠AOB ii) ∠AOD iii) ∠BOE iv) ∠AED v) ∠BDE
vi) ∠BOC vii) ∠BOF viii) ∠AOF ix) ∠COD
x) Find∠DOE by using the result of∠AOB
xi) Find∠FOE by using the result of ∠BOF and ∠DOE.
xii) By using the result of∠AOB, ∠AOD, ∠BOF and ∠DOE, find∠FOH if ∠HOE=20°. (adj.∠s on st. line is NOT allowed to use)


Solution:
i) ∠AOB = 180°-55°-60° (∠sum of△) = 65°
ii) ∠AOD = 55°+60° (ext.∠of△) = 115°
iii) ∠BOE = 55°+60° (ext.∠of△) = 115°
iv) ∠AED = ∠BAO (alt.∠s, BA//ED) = 60°
v) ∠BDE = ∠ABO (alt.∠s, BA//ED) = 55°
vi) ∠BOC = 180°-55° (int.∠s, BA//CF) = 125°
vii) ∠BOF = ∠ABO (alt.∠s, BA//CF) = 55°
viii) ∠AOF = 180°-60° (int.∠s, BA//CF) = 120°
ix) ∠COD =∠ABO (corr.∠s, BA//CF) = 55°
x) ∠DOE = ∠AOB(vert. opp.∠s) = 65° (the result from i)
xi) ∠FOE = 180°-∠BOF-∠DOE (adj.∠s on st. line)
= 180°- 55° - 65° = 60°
xii) ∠FOH = 360°–∠AOB –∠AOD –∠BOF –∠DOE –∠HOE (∠s at a pt.)
= 360°– 65°–115°–55°–65°–20° = 40°

你好！ 我建議你花幾十元到書局親自選購一本補充練習。 在商務書局見過一本橙色、英文的數學補充，本子小小但題目齊全，可惜被媽媽弄掉了，不能提供相關書名、出版社。有時間可到各大書局找找適合自己的。

打中文啦!!