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數學知識交流 － 幾何
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA01076848/o/701204230069213873398430.jpg AB = BC = CA = AD，求 ∠BDC。
最佳解答:
設∠ABD = ∠ADB = k 則∠DBC = 60° - k ∠ACD = ∠ADC = k +∠BDC △BCD內角和 = ∠DBC + ∠ACB + ∠ACD + ∠BDC = 180° 60° - k + 60° + k +∠BDC + ∠BDC = 180° ∠BDC = 30°
I also thought that. But this is just a F.1 questions (taken from http://www.hcm.org.hk/index.html), do we have a easier to do it (not easy, simple)?|||||30 (by a bit circle geometry)
數學知識交流 － 幾何
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA01076848/o/701204230069213873398430.jpg AB = BC = CA = AD，求 ∠BDC。
最佳解答:
設∠ABD = ∠ADB = k 則∠DBC = 60° - k ∠ACD = ∠ADC = k +∠BDC △BCD內角和 = ∠DBC + ∠ACB + ∠ACD + ∠BDC = 180° 60° - k + 60° + k +∠BDC + ∠BDC = 180° ∠BDC = 30°
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其他解答:I also thought that. But this is just a F.1 questions (taken from http://www.hcm.org.hk/index.html), do we have a easier to do it (not easy, simple)?|||||30 (by a bit circle geometry)
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