#include <iostream>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <stdio.h>
 
using namespace std;
 
//-------------------------------------------------------------------------
// geometric library
 
namespace geometry {
 
#define EPS 1E-10 ///< 0とみなすサイズ。適当

struct point {
    double x;
    double y;

	point& operator += (const point& rhs) { x += rhs.x; y += rhs.y; return *this; }
	point& operator -= (const point& rhs) { x -= rhs.x; y -= rhs.y; return *this; }
	point& operator /= (double factor) { x /= factor; y /= factor; return *this; }
	point& operator *= (double factor) { x *= factor; y *= factor; return *this; }
	bool operator == (const point& rhs) const { return (std::abs(x - rhs.x) < EPS) && (std::abs(y - rhs.y) < EPS); }
	bool operator != (const point& rhs) const { return !(*this == rhs); }
};

typedef point vector;
 
struct circle {
    point center;
    double radius;
};
 
struct line { // ax + by + c = 0
    double a;
    double b;
    double c;
};

enum rotate_dir {
	rotate_None = 0,
	rotate_CW   = 1,
	rotate_CCW  = -1
};

point operator + (const point& p1, const point& p2)
{
	point p(p1);
	p += p2;
	return p;
}

point operator / (const point& p1, double factor)
{
	point p(p1);
	p /= factor;
	return p;
}
 
point operator - (const point& p1, const point& p2)
{
	point p(p1);
	p -= p2;
	return p;
}

std::istream& operator >> (std::istream& is, point& p)
{
    is >> p.x >> p.y;
    return is;
}

///! 2点の距離
double get_distance(const point& p1, const point& p2)
{
    return sqrt(pow(p1.x - p2.x, 2) + pow(p1.y - p2.y, 2));
}
 
///! 2直線の交点
bool get_crosspoint(const line& l1, const line& l2, point *result)
{
    double denominator = l1.a * l2.b - l1.b * l2.a;
    if (denominator == 0) {
        return false;
    }
    result->x = (l1.b * l2.c - l2.b * l1.c) / denominator;
    result->y = (l2.a * l1.c - l1.a * l2.c) / denominator;
    return true;
}
 
///! 2点の垂直2等分線
bool get_bisector(const point& p1, const point& p2, line *result)
{
    if (p1 == p2) {
        return false;
    }
    result->a = 2 * (p1.x - p2.x);
    result->b = 2 * (p1.y - p2.y);
    result->c = -(p1.y - p2.y) * (p1.y + p2.y) - (p1.x - p2.x) * (p1.x + p2.x);
    return true;
}
 
///! 外接円を求める
bool get_circum_circle(const point& p1, const point& p2, const point& p3, circle *result)
{
    double a = get_distance(p1, p2);
    double b = get_distance(p2, p3);
    double c = get_distance(p3, p1);
 
    // 半径の公式
    result->radius = a * b * c / sqrt((a + b + c) * (-a + b + c) * (a - b + c) * (a + b - c));
 
    line line_p1p2;
    line line_p2p3;
 
    // 頂点の垂直2等分線の交点が外心
    return get_bisector(p1, p2, &line_p1p2) &&
           get_bisector(p2, p3, &line_p2p3) &&
           get_crosspoint(line_p1p2, line_p2p3, &result->center);
}

double cross_product(const point& p1, const point& p2)
{
	return p1.x * p2.y - p2.x * p1.y;
}

///! p1-p2-p3が左右どちらの方向に折れているかを判定
rotate_dir direction(const point& p1, const point& p2, const point& p3)
{
	vector p1_p3 = p1 - p3;
	vector p2_p3 = p2 - p3;
	double cross = cross_product(p1_p3, p2_p3);

	if (std::abs(cross) < EPS) {
		return rotate_None;
	} else if (cross > 0) {
		return rotate_CW;
	} else {
		return rotate_CCW;
	}
}
 
// (p1s, p1e)と(p2s, p2e)の線分交差判定.EPS以下で接している場合はtrue、重なっている場合はfalse
bool intersect(const point& p1s, const point& p1e, const point& p2s, const point& p2e)
{
	assert(p1s != p1e && p2s != p2e);
	return 
		(direction(p1s, p1e, p2s) != direction(p1s, p1e, p2e)) &&
		(direction(p2s, p2e, p1s) != direction(p2s, p2e, p1e));
}

// targetが三角形p1,p2,p3の内部にあるかどうか
bool is_inside_triangle(const point& p1, const point& p2, const point& p3, const point& target)
{
	point center_of_gravity = (p1 + p2 + p3) / 3; // 重心

	if (center_of_gravity == target) return true;

	// 重心とtargetを結んだ線分が三角形の辺と交叉するなら、点は三角形の外部にある
	bool intersect_any =
		intersect(p1, p2, center_of_gravity, target) ||
		intersect(p2, p3, center_of_gravity, target) ||
		intersect(p3, p1, center_of_gravity, target);

	return !intersect_any;
}

} // namespace geometry
 
//-------------------------------------------------------------------------
// solution
 
using namespace geometry;

int main(void)
{
	point p1,p2,p3,target;

	while (cin >> p1 >> p2 >> p3 >> target) {
		cout << (is_inside_triangle(p1, p2, p3, target) ? "YES\n" : "NO\n");
	}

	return 0;
}

Measured memory_bytes: 1192