If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle

If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle

The original formula can be changed into 2A2 + 2B2 + 2c2 = 2Ab + 2Ac + 2BC, that is, A2 + B2 + C2 + A2 + B2 + c2-2ab-2ac-2bc = 0;
According to the complete square formula, we get: (a-b) 2 + (C-A) 2 + (B-C) 2 = 0;
From the properties of nonnegative numbers, we can know that A-B = 0, C-A = 0, B-C = 0; that is, a = b = C. Therefore, △ ABC is an equilateral triangle
Therefore, C

The triangle triangle a.b.c. satisfies that a square plus b square plus C square equals AB plus BC plus AC. try to judge the shape of the triangle ABC

A square + b square + C square = AB + BC + Ca
A square + b square + C square - (AB + BC + Ca)
=1/2*[(a-b)^2+(b-c)^2+(c-a)^2]
=0
a-b=0,b-c=0,c-a=0
a=b,b=c,c=a
So:
a=b=c
The triangle is equilateral

2. If A.B.C is the three sides of the triangle ABC, and the square of a + the square of B + the square of C = AB + AC + BC

Let the square of a + the square of B + the square of C = AB + AC + BC, let the two sides of the equation multiply by 2 to obtain 2 (the square of a + the square of B + the square of C) = 2Ab + 2Ac + 2BC, that is, 2 (the square of a + the square of B + the square of C) - 2Ab + 2Ac + 2BC = 0, (the square of a + the square of B - 2BC) + (the square of a + the square of C - 2Ac)

If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle

The original formula can be changed into 2A2 + 2B2 + 2c2 = 2Ab + 2Ac + 2BC, that is, A2 + B2 + C2 + A2 + B2 + c2-2ab-2ac-2bc = 0;
According to the complete square formula, we get: (a-b) 2 + (C-A) 2 + (B-C) 2 = 0;
From the properties of nonnegative numbers, we can know that A-B = 0, C-A = 0, B-C = 0; that is, a = b = C. Therefore, △ ABC is an equilateral triangle
Therefore, C

It is known that a, B, C are the three sides of the triangle ABC, and the square of a + the square of bc-ac-b = 0

a^2+bc-ac-b^2=0
a^2-b^2-c(a-b)=0
(a+b)(a-b)-c(a-b)=0
(a-b)(a+b-c)=0
A-B = 0 a + B-C = 0
A=b
So it's an isosceles triangle

The triangle ABC triangle A.B.C is known to satisfy the square of a + the square of B + the square of C = AB + BC + AC

Equilateral triangle
a^2+b^2+c^2=ab+ac+bc
2a^2+2b^2+2c^2=2ab+2ac+2bc
2a^2+2b^2+2c^2-2ab-2ac-2bc=0
(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
(a-b)^2+(a-c)^2+(b-c)^2=0
∴a-b=0 a-c=0 b-c=0
∴a=b a=c b=c
That is, a = b = C
Get an equilateral triangle

It is known that a, B and C are the three sides of △ ABC and satisfy A2-B2 + AC BC = 0. Please judge the shape of △ ABC

a2-b2+ac-bc=0，
From the formula of square difference, we can get that:
（a+b）（a-b）+c（a-b）=0，
（a-b）（a+b+c）=0，
∵ a, B and C are the sides of a triangle,
ν a, B and C are all greater than 0,
The solution of this equation is a = B,
The △ ABC must be an isosceles triangle

If the triangle ABC of the triangle ABC satisfies the square of a + the square of B + the square of C = AB + BC + Ca, the shape of the triangle ABC is judged

a²+b²+c²=ab+bc+ca
a²+b²+c²-ab-bc-ac=0
Double 2 on both sides
2a²+2b²+2c²-2ab-2bc-2ac=0
(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)=0
(a-b)²+(b-c)²+(c-a)²=0
If the square is greater than or equal to 0, the sum is equal to 0. If there is one greater than 0, at least one is less than 0, which is not true. So all three are equal to 0
So A-B = 0, B-C = 0, C-A = 0
a=b,b=c,c=a
So a = b = C
So it's an equilateral triangle

If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle

The original formula can be changed into 2A2 + 2B2 + 2c2 = 2Ab + 2Ac + 2BC, that is, A2 + B2 + C2 + A2 + B2 + c2-2ab-2ac-2bc = 0;
According to the complete square formula, we get: (a-b) 2 + (C-A) 2 + (B-C) 2 = 0;
From the properties of nonnegative numbers, we can know that A-B = 0, C-A = 0, B-C = 0; that is, a = b = C. Therefore, △ ABC is an equilateral triangle
Therefore, C

If a, B, C are the three sides of the triangle ABC, and the square of a + the square of B + the square of C - AB BC CA = 0, explore the shape of triangle ABC and write the reason

It is because that a 2 + B 2 + C 2 - AB BC CA = 0
So 2A 2 + 2B 2 + 2C 2 - 2ab-2bc-2ca = 0
(a²-2ab+b²)+(a²-2ca+c²)+(b²-2bc+c²)=0
(a-b)²+(a-c)²+(b-c)²=0
So A-B = 0 and a-c = 0, and B-C = 0
a=b,a=c,b=c
So a = b = C
Therefore, the triangle ABC is an equilateral triangle