標題:
中二數學(Angles in Triangles and Polygons)
發問:
Exploring Mathematics 2B Ch.8 P.21 第23題唔識做, 請指教. 題目是:In 三角形ABC, AB=BC. The angle bisector of 角ABC is drawn to meet AC at D. A line from D is drawn to meet BC at E such that AD=DE. If DE//AB, find 角 DCE. 更新: 係要揾角DCE, 答案係30度, 但我唔識揾. 麻煩哂各位! 更新 2: 我條係8B第23題, 本書底答案明明寫係30度.
最佳解答:
我都係用e本書...但係係中文...我盡量用英文la~ 仲有...角 DCE=60度 唔係30度 睇清楚D AD=DC (因為係平分線,所以AD=DC)<---唔洗寫 BE=CE (你係AE條線+一條平分線) AD=DE (given) 所以DC=DE=EC 角EDC=角DCE=角DEC=60度(等邊三角形性質) 2008-03-26 14:43:11 補充： 你果條係8B第23條
Yes, 要揾角DCE. 答案係30度, 但我唔知點揾. Anyway, 唔該哂!|||||are you sure that you are asked to find 角 DCE, not CDE or DEC? because, if this is the case, all those lines drawing and infomation of AD=DE are very unneccessary! Anyway, 角DCE = 角ACB due to the question provided DE//AB, then Triangle DEC is a smaller version of Triangle ABC. thus, all angles are the same
中二數學(Angles in Triangles and Polygons)
發問:
Exploring Mathematics 2B Ch.8 P.21 第23題唔識做, 請指教. 題目是:In 三角形ABC, AB=BC. The angle bisector of 角ABC is drawn to meet AC at D. A line from D is drawn to meet BC at E such that AD=DE. If DE//AB, find 角 DCE. 更新: 係要揾角DCE, 答案係30度, 但我唔識揾. 麻煩哂各位! 更新 2: 我條係8B第23題, 本書底答案明明寫係30度.
最佳解答:
我都係用e本書...但係係中文...我盡量用英文la~ 仲有...角 DCE=60度 唔係30度 睇清楚D AD=DC (因為係平分線,所以AD=DC)<---唔洗寫 BE=CE (你係AE條線+一條平分線) AD=DE (given) 所以DC=DE=EC 角EDC=角DCE=角DEC=60度(等邊三角形性質) 2008-03-26 14:43:11 補充： 你果條係8B第23條
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其他解答:Yes, 要揾角DCE. 答案係30度, 但我唔知點揾. Anyway, 唔該哂!|||||are you sure that you are asked to find 角 DCE, not CDE or DEC? because, if this is the case, all those lines drawing and infomation of AD=DE are very unneccessary! Anyway, 角DCE = 角ACB due to the question provided DE//AB, then Triangle DEC is a smaller version of Triangle ABC. thus, all angles are the same
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