已知C是AB上一点,AC＝根号5-1,请你证明AC＝AB*BC

第一题,已知C是AB上一点,AB＝2,AC＝根号5-1,请你证明AC＝AB*BCAB = 2AC = √5 - 1BC = 2 - (√5 - 1) = 3 - √5 左端 = AC² = (√5 - 1)² = 5 - 2√5 +1 = 6 -2√5 右端 = AB*BC = 2 * (3 -√5 ) = 6 -2√5左端 = AC² = AB*BC = 右端第二题,已知AB＝2,点C是AB的黄金分割点,点D在AB上,且AD的平方＝BD*AB,求CD除以AC的值（如图）设 AD = xAD² = BD*AB ===> x² = (2 - x)*2 = 4 - 2x ===> x = ±√5 - 1 (舍负) = √5 - 1可见D点和C点重合CD = 0是不是题目给错了：BD的平方＝AD*AB 如果是这样的话：BD = √5 - 1 AD = 3 - √5CD = AC - AD = (√5 - 1) - (3 - √5) = 2√5 - 4AC = √5 - 1CD / AC = 2(√5 - 2) / (5 - 1) = (3 - √5) / 2注释：(√5 - 1) / 2 ≈0。618 黄金分割数,没那么神秘,只不过是一个数,有人吹捧而已(√5 + 1) / 2 ≈1。618(3 - √5) / 2 ≈ 1 - 0。618 = 0。382