def is_opposit(p0,p1,p2,p3):
    x0,y0 = p0; x1,y1 = p1; x2,y2 = p2; x3,y3 = p3
    if ((x0-x1)*(y2-y0)+(y0-y1)*(x0-x2))*((x0-x1)*(y3-y0)+(y0-y1)*(x0-x3)) >= 0:
        return False
    else:
        return True

# 線分 p0--p1 と 線分 p2--p3 は交わるか？
# ただし一般の位置を仮定している
def is_crossed(p0,p1,p2,p3):
    return is_opposit(p0,p1,p2,p3) and is_opposit(p2,p3,p0,p1)

def is_star(a):
    for i in range(5):
        if not is_crossed(a[i],a[(i+2)%5],a[(i+1)%5],a[(i+3)%5]): return False
    return True

xy = [list(map(int,input().split())) for _ in range(5)]
from itertools import permutations
for a in permutations(xy):
    if is_star(a):
        print("YES")
        exit()
print("NO")