是否存在一個非直角三角形ABC,它的三條邊滿足AB² + BC² = CA² ?如何證明?

use cos theory, cosc=(AB^2+BC^2-CA^2)/(2ab)
if angle A,B &C are<90,
cosa, cosb, cosc>0
if cosA,cosB &cosC>0, base on cos theory:
cosc=(AB^2+BC^2-CA^2)/(2ab)
only right triangle has AB^2+BC^2=CA^2
you will find that AB^2+BC^2is not equal to CA^2