Даны четыре точки на плоскости, определить какое расстояние меньше

#include <iostream>
#include <cmath>
 
using namespace std;
 
int main(){
    
    double ax, ay;
    cout << "a: "; cin >> ax >> ay; 
    double bx, by;
    cout << "b: "; cin >> bx >> by; 
    double cx, cy;
    cout << "c: "; cin >> cx >> cy; 
    double dx, dy;
    cout << "d: "; cin >> dx >> dy;
    
    double ab = sqrt((ax - bx)*(ax - bx) + (ay - by)*(ay - by));
    double bc = sqrt((cx - bx)*(cx - bx) + (cy - by)*(cy - by));
    double ac = sqrt((ax - cx)*(ax - cx) + (ay - cy)*(ay - cy));
    double cd = sqrt((cx - dx)*(cx - dx) + (cy - dy)*(cy - dy));
    
    if (ab + bc < ac + cd)
        cout << "ab + bc < ac + cd" << endl;
    else
        cout << "ab + bc >= ac + cd" << endl;
}

#include <iostream>
#include <math.h>
 
using namespace std;
 
struct  Point
{
    int x;
    int y;
};
 
 
int main(int argc, char const *argv[])
{
    Point A, B, C, D;
    cout << "Enter A(x)= "; cin >> A.x; cout << "Enter A(y)= "; cin >> A.y;
    cout << "Enter B(x)= "; cin >> B.x; cout << "Enter B(y)= "; cin >> B.y;
    cout << "Enter C(x)= "; cin >> C.x; cout << "Enter C(y)= "; cin >> C.y;
    cout << "Enter D(x)= "; cin >> D.x; cout << "Enter D(y)= "; cin >> D.y;
 
    double AB = sqrt((B.x - A.x) ^ 2 + (B.y - A.y) ^ 2);
    double BC = sqrt((C.x - B.x) ^ 2 + (C.y - B.y) ^ 2);
    double AC = sqrt((C.x - A.x) ^ 2 + (C.y - A.y) ^ 2);
    double CD = sqrt((D.x - C.x) ^ 2 + (D.y - C.y) ^ 2);
 
    if ((AB + BC) < (AC + CD))
        cout << "AB + BC less than AC + CD" << endl;
    else if ((AB + BC) == (AC + CD))
        cout << "AB + BC equals AC + CD" << endl;
    else
        cout << "AC + CD less than AB + BC" << endl;
    return 0;
}