The equations x−y=4 and x2+4xy+y2=0 represented by the side of

Acute angle between the lines x2+4xy+y2=0 is tan−1{(24−1)/(1+1)}=tan−13=π/3. The angle bisectors of x2+4xy+y2=0  are given by x2−y21−1=xy or x2−y2=0or x=±yAs x+y=0 is perpendicular to x−y=4,the given triangle is isosceles with vertical angle equal to π/3.Hence, it is an equilateral triangle.