本區搜索:
Yahoo!字典
打印

[Core] 幾何難題

[隱藏]

幾何難題

怎樣在不計算長度而求到x和a?
   

TOP

回覆 1# cindylee2013 的帖子

cannot see the picture clearly

TOP

回覆 1# cindylee2013 的帖子



1. Calculate some known angles:
ACB = 180-(10+70)-(60+20) = 20°
AEB = 180-70-(60+20) = 30°

2. Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude:
DCF  ACB
CFD = CBA = 60+20 = 80°
DFB = 180-80 = 100°
CDF = CAB = 70+10 = 80°
ADF = 180-80 = 100°
BDF = 180-100-20 = 60°

3. Draw a line FA labeling the intersection with DB as a new point G and conclude:
ADF  BFD
AFD = BDF = 60°
DGF = 180-60-60 = 60° = AGB
GAB = 180-60-60 = 60°
DFG (with all angles 60°) is equilateral
AGB (with all angles 60°) is equilateral
4. CFA with two 20° angles is isosceles, so FC = FA

5. Draw a line CG, which bisects ACB and conclude:
ACG  CAE
FC-CE = FA-AG = FE = FG
FG = FD, so FE = FD

6. With two equal sides, DFE is isosceles and conclude:
DEF = 30+x = (180-80)/2 = 50
Answer: x = 20°

TOP

重要聲明:小卒資訊論壇 是一個公開的學術交流及分享平台。 論壇內所有檔案及內容 都只可作學術交流之用,絕不能用商業用途。 所有會員均須對自己所發表的言論而引起的法律責任負責(包括上傳檔案或連結), 本壇並不擔保該等資料之準確性及可靠性,且概不會就因有關資料之任何不確或遺漏而引致之任何損失或 損害承擔任何責任(不論是否與侵權行為、訂立契約或其他方面有關 ) 。