python:任意输入3个数,判断能否组成三角形

任意输入3个数,判断能否组成三角形,并输出三角形为等边/等腰/直角/普通三角形.
三角形:两边之和大于第三边
直角三角形:勾股定理

#!/usr/bin/python
# -*- coding:utf-8 -*-
#输入合法性检查,必须输入正数,不支持科学计数法
def ispositive(numb):
    try:
        float(numb)
    except:
        return False
    else:
        if float(numb) <= 0:
            return False
        else:
            return True
#直角三角形判断
def ispythagoras(a,b,c):
    if a**2 + b**2 == c**2 or a**2 + c**2 == b**2 or b**2 + c**2 == a**2:
        return True
    else:
        return False

num1 = input("pls enter 1st number:\n")
while not ispositive(num1):
    num1 = input("That's not a valid number. Try again:\n")

num2 = input("pls enter 2nd number:\n")
while not ispositive(num2):
    num2 = input("That's not a valid number. Try again:\n")

num3 = input("pls enter 3rd number:\n")
while not ispositive(num3):
    num3 = input("That's not a valid number. Try again:\n")

num1 = float(num1)
num2 = float(num2)
num3 = float(num3)
#欢迎点评,引用请注明出处
if num1 + num2 > num3 and num2 + num3 > num1 and num1 + num3 > num2:
    if num1 == num2 == num3:
        print("%.2f\n%.2f\n%.2f\n可以组成等边三角形" % (num1,num2,num3))
    elif num1 == num2 or num2 == num3 or num1 == num3:
        if ispythagoras(num1,num2,num3):
            print('%.2f\n%.2f\n%.2f\n可以组成等腰直角三角形' % (num1,num2,num3))
        else:
            print('%.2f\n%.2f\n%.2f\n可以组成等腰三角形' % (num1,num2,num3))
    elif ispythagoras(num1,num2,num3):
        print('%.2f\n%.2f\n%.2f\n可以组成直角三角形' % (num1,num2,num3))
    else:
        print('%.2f\n%.2f\n%.2f\n可以组成普通三角形' % (num1,num2,num3))
else:
    print('%.2f\n%.2f\n%.2f\n不能组成三角形' % (num1,num2,num3))