In the triangle ABC, if (AB) ^ 2 + (AB) * (BC) = 0, then the shape of the triangle is () is a vector Explain

In the triangle ABC, if (AB) ^ 2 + (AB) * (BC) = 0, then the shape of the triangle is () is a vector Explain

（AB）^2＋（AB）＊（BC）＝0,
AB^2+AB*BC*cos(π-B)=0
AB^2-AB*BC*cosB=0
AB-BC*cosB=0
cosB=(AB^2+BC^2-AC^2)/2AB*BC
That is ab = BC * CoSb = BC * (AB ^ 2 + BC ^ 2-ac ^ 2) / (2Ab * BC)
2AB^2=AB^2+BC^2-AC^2
So AB ^ 2 + AC ^ 2 = BC ^ 2
So a triangle is a right triangle

Ad is the middle line of triangle ABC, e is the extension line of a point be CE on ad, which respectively intersects AC AB at point Mn to verify Mn / / BC

[sorry, it's too late to see the title] this problem can be proved by the area method. (an important theorem to be used in this problem is: the area ratio of two triangles of the same height is equal to the ratio of the bottom edge) the proof is: ∵ △ AEC and △ Dec are of the same height ∵ s △ AEC: s △ Dec = AE: ED, s △ AEB: s △ bed = AE: ed ∵ s △ AEC: s △

It is known that A.B.C is the length of three sides of △ ABC, and satisfies a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac = 0. Please judge the shape of △ ABC

Two sides of the equation x2
（a²+b²-2ab）+（a²+c²-2ac）+（b²+c²-2bc）=0
（a-b）²+（a-c）²+（b-c）²=0
therefore
a-b=0
a-c=0
b-c=0
therefore
a=b=c
So △ ABC is an equilateral triangle

Given that a, B and C are three sides of triangle ABC, and satisfy a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac = 0, try to judge its shape

Equilateral triangle
a^2+b^2+c^2-ab-bc-ac＝0
Well organized
1/2[(a-b)^2+(b-c)^2+(a-c)^2]=0
So a = B, B = C, a = C
So a = b = C
So equilateral triangles