✔ 最佳答案
.............A_______B
............../..........
............./.......
............/....
........D/__________ C
設 ㄥ BDC = x ,
則ㄥ ABD = x (alt ㄥs , AB // CD)
故ㄥADB = 180° - 2x考慮△DBC , 餘弦定理 :cos x
= (a² + a² - b²) / (2aa)
= (2a² - b²) / (2a²)考慮△DAC , 餘弦定理 :AC² = a² + a² - 2 a a cos (180° - 2x + x)
AC² = 2a² (1 - cos(180° - x))
AC² = 2a² (1 + cos x)
AC² = 2a² (1 + (2a² - b²) / (2a²))
AC² = 4a² - b²
AC = √ [ (2a - b) (2a + b) ]
2011-04-27 21:27:26 補充:
另解 :
以D為圓心 , a 為半徑作圓 , 延長 BD 交圓於 E ,
則 ㄥCDE = 180° - x = ㄥADC ,
故等腰△DAC ≡ 等腰△DCE
故 CE = CA ,
又 △ ECB 是直角△ , 得
CA = CE = √ (EB ² - BC ²) = √ ( (2a)² - b² ) = √ [ (2a - b) (2a + b) ]