It is known that a, B, C are the three sides of △ ABC and satisfy the relation of a 2 + C 2 = 2 ab + 2 BC-2 B 2. Try to judge the shape of a triangle

It is known that a, B, C are the three sides of △ ABC and satisfy the relation of a 2 + C 2 = 2 ab + 2 BC-2 B 2. Try to judge the shape of a triangle

Therefore, if A-B = 0 and B-C = 0, then a = b = C triangle ABC is 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

It is known that a, B and C are the three sides of △ ABC and satisfy the relation a  2 + C  = 2Ab + 2ac-2b 2. Try to judge the shape of △ ABC Detailed process

∵a²+c²=2ab+2bc-2b²
a²+c²-2ab-2bc+2b²=0
a²-2ab+b²+b²-2bc+c²=0
(a-b)²+(b-c)²=0
∴a-b=0 b-c=0
a=b=c
The shape of △ ABC is an equilateral triangle

It is known that the side length of △ ABC is a, B, C, and a  2 B  2 + C  - 2 ab-2bc = 0, try to judge the shape of △ ABC

Because the formula a 2 + 2 B 2 + C 2 a B-2 B C = 0, a 2 + B 2 a B + B 2 + C 2 - 2 BC = 0, that is (a-b) ^ 2 + (B-C) ^ 2 = 0, because the square number of a number is greater than or equal to 0, and now the sum of the two is 0, so they are 0 respectively, that is, A-B = 0, B-C = 0, that is, a = B, B = C, so a = b = C

It is known that ABC is the length of the three sides of the triangle ABC, and satisfies the requirements of a 2 + 2 B 2 + C 2 B (a + C) = 0, try to judge the shape of the triangle 1: It is known that ABC is the length of three sides of the triangle ABC, and satisfies the requirements of a 2 + 2 B 2 + C 2 B (a + C) = 0? 2: Given that XY is any rational number, M = x 2 + y 2, n = 2XY, can you determine the size of M and N? Why?

1: It is known that ABC is the length of three sides of the triangle ABC, and satisfies the requirements of a 2 + 2 B 2 + C 2 B (a + C) = 0?
The reason is that a 2 + 2 B 2 + C 2 B (a + C) = 0
That is, a 2 + 2 B 2 + C 2 - 2 ab - 2 BC = 0
(a²-2ab+b²)+(b²-2bc+c²)=0
(a-b）²+(b-c)²=0
So: A-B = 0, B-C = 0
So a = b = C
So a triangle is an equilateral triangle
2: Given that XY is any rational number, M = x? + y?, n = 2XY, can you determine the size of M and N? Why?
m-n=x²+y²-2xy
=（x-y）²≥0
So m ≥ n

If a, B, C of the triangle ABC satisfy the relation | B-12 | + | C-13 | + (a + 2b-c-16) ^ 2 = 0, then a =? B =? C =?

|b-12|+|c-13|+(a+2b-c-16)^2=0
Because | B-12 | > = 0, | C-13 | > = 0, (a + 2b-c-16) ^ 2 > = 0
And their sum is zero,
So it can only be:
Each is equal to 0
therefore
b-12=0
c-13=0
a+2b-c-16=0
b=12,
c=13
a=22

It is known that a, B, C are the three side lengths of △ ABC, and satisfy the relationship a  2 + C  = 2Ab + 2bc-2b 2. It is shown that △ ABC is a triangle

Wait

If the three sides of △ ABC satisfy a2-2bc = c2-2ab, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Acute triangle

The equation can be transformed into: a2-2bc-c2 + 2Ab = 0,
（a2-c2）+（2ab-2bc）=0，
（a+c）（a-c）+2b（a-c）=0，
（a-c）（a+c+2b）=0，
∵ a, B, C are the three sides of ∵ ABC,
∴a+c+2b＞0，
∴a-c=0，
∴a=c．
The triangle is an isosceles triangle,
Therefore, a

If the three sides a, B and C of △ ABC have the following relation: a  C  2 + 2ab-2cb = 0, it indicates that the triangle form isosceles triangle

A²-C²+2AB-2CB=0
A²-C²+2AB-2CB+B²-B²=0
(A+B)²-(B+C)²=0
(A+B)²=(B+C)²
A+B = B+C
A=C
So it's isosceles

Let a, B, C be the three sides of △ ABC, and the square of a-2bc = the square-2ac of B, judge the shape of triangle ABC

There are two types of a-2bc = B? - 2Ac: a? B? 2 + 2ac-2bc = 0
∴﹙a-b﹚﹙a+b﹚+2c﹙a-b﹚=0
﹙a-b﹚﹙a+b+2c﹚=0
A-B = 0 or a + B + 2C = 0 ∵ a, B, C are the three sides of a triangle, so a + B + 2C = 0 does not hold
 there is: A-B = 0  a = B
The triangle is an isosceles triangle