It is known that a, B, C are the lengths of the three sides of the triangle ABC, and the square of a + 2B + C - 2b (a + C) = 0, try to judge the shape of the triangle Be clear,

It is known that a, B, C are the lengths of the three sides of the triangle ABC, and the square of a + 2B + C - 2b (a + C) = 0, try to judge the shape of the triangle Be clear,

Is the title wrong? It should be + 2B ^ 2, not + 2B
The answer is an equilateral triangle

Given that a, B, C are the three sides of the triangle ABC and satisfy the square of a + the square of C = 2Ab + 2bc-2b, it is proved that the triangle ABC is an equilateral triangle

(a²-2ab+b²)+(b²-2bc+c²)=0
(a-b)²+(b-c)²=0
Then A-B = B-C = 0
a=b,b=c
So a = b = C
So it's an equilateral triangle

It is known that a, B, C are the three sides of △ ABC and satisfy the relation A2 + C2 = 2Ab + 2bc-2b2. It is shown that △ ABC is an equilateral triangle

∵ the original formula can be changed into A2 + c2-2ab-2bc + 2B2 = 0,
a2+b2-2ab+c2-2bc+b2=0，
That is (a-b) 2 + (B-C) 2 = 0,
ν A-B = 0 and B-C = 0, that is, a = B and B = C,
∴a=b=c．
So △ ABC is an equilateral triangle

If the three sides of a triangle a, B, C satisfy a square + 2B square + C square - 2ab-2bc = O, try to explain that the triangle ABC is equilateral

A square + 2B square + C square - 2ab-2bc = O
A square + b square - 2Ab + b square + C square - 2BC = O
(a-b) square + (B-C) square = 0
So a = B, B = C;
So a = b = C, that is, the triangle ABC is equilateral

It is known that a, B and C are the three sides of the triangle ABC and satisfy the quadratic power of a + the second power of C = the square of 2Ab + 2bc-2b. Try to judge what shape the triangle ABC is and explain the reasons

a^2 + c^2 = 2ab + 2bc - 2b^2
a^2 - 2ab + b^2 + c^2 - 2bc + b^2 = 0
(a-b)^2 + (c-b)^2 = 0
∵ (a-b) ^ 2 constant ≥ 0, (C-B) ^ 2 constant ≥ 0
∴a-b=0 ,c-b=0
That is: a = B, C = B
∴a=b=c
Triangle ABC is an equilateral triangle

It is known that a, B, C are the lengths of the three sides of △ ABC and satisfy a ^ 2 + 2B ^ 2 + C ^ 2-2b (a + C) = O. try to judge the shape of this triangle

because
a^2+2b^2+c^2－2b(a+c)=0
a^2-2ab+b^2+b^2-2bc+c^2=0
(a-b)^2+(b-c)^2=0
therefore
a-b=0
A=b
b-c=0
B=c
So a = b = C triangle is an equilateral triangle

Let a, B, C be the length of the three sides of △ ABC and satisfy a ^ 2 + 2B ^ 2 + C ^ 2-2b (a + b) = 0, try to judge the shape of this triangle Then the value of ^ 2-0b is ^ 2-0b____

a^2+2b^2+c^2-2b(a+c)
=(a^2+b^2-2ab)+(b^2+c^2-2bc)
=(a-b)^2+(b-c)^2=0
So: a = B, B = C
That is, a = b = C
The triangle is equilateral

If a, B and C are the lengths of the three sides of △ ABC and satisfy A2 + 2B2 + c2-2b (a + C) = 0, then the shape of the triangle is______ .

From the known condition A2 + 2B2 + c2-2b (a + C) = 0, it is obtained that,
（a-b）2+（b-c）2=0
∴a-b=0，b-c=0
That is, a = B, B = C
∴a=b=c
So the answer is an equilateral triangle

It is known that in △ ABC, if the three side lengths a, B, C satisfy the equation a2-16b2-c2 + 6ab + 10bc = 0, then () A. a+c＞2b B. a+c=2b C. a+c＜2b D. The relationship between a + C and 2b is uncertain

(a + 3b) 2 - (c-5b) 2 = 0, (a + 3B + c-5b) (a + 3B + c-5bb) (a + 3B + c-5bb) (a + 3B + c-5b) (a + 3B + c-5b) (a + 3b-c + 5b) (a + 3b-c + 5b) = 0, that is (a + c-2b) (A-C + 8b) = 0, namely (a + c-2b) (A-C + 8b) = 0, a + c-2b = 0 or a-c + 8b = 0, \a + C = 2B or a + 8b = C, M a + b > C, \\\\\\\\\b = C does not conform to the meaning of the question

In △ ABC, if a, B, C satisfy (a + B + C) (a + B-C) = 3AB, then ∠ C=______ .

∵（a+b+c）（a+b-c）=3ab，
∴（a+b）2-c2=3ab
∴a2+b2-c2=ab
According to the cosine theorem, the following results are obtained
cosC=a2+b2−c2
2ab=1
Two
C=60°
So the answer is: 60 degrees