✔ 最佳答案
其實唔難
諗下直角三角形同畢氏定理有關之後再計
設 ( 可設可唔設,不過方便而已 )
AC = a
AB = b
BC = c
2a = b+c -------------- 1
a^2 + b^2 = c^2 (畢氏定理) --------------- 2
從1 ,
c = 2a - b ------------ 3
從2 ,
c = √(a^2 + b^2) -------- 4
3 代入 4
2a - b = √(a^2 + b^2)
(2a - b)^2 = (a^2 + b^2)
4a^2 - 2(2a)(b) + b^2 = a^2 + b^2 解:∵ (a - b)^2 ≡ a^2 -2(a)(b) + b^2
4a^2 - 4ab + b^2 = a^2 + b^2
3a^2 - 4ab = 0
3a^2 = 4ab
a^2 / ab = 3 / 4
a / b = 3 / 4
a : b = 3 : 4
根據畢氏定理,
a^2 + b^2 = c^2
3^2 + 4^2 = c^2
9 + 16 = c^2
25 = c^2
c = 5
∴ a:b:c = 3:4:5 ( AC:AB:BC = 3:4:5 )