Given that the triangles a, B, C satisfy the equation a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac = 0, try to judge the shape of △ ABC There should be an analysis process

a^2+b^2+c^2-ab-bc-ac=0
Left = [(a ^ 2 + B ^ 2-2ab) + (a ^ 2 + C ^ 2-2ac) + (b ^ 2 + C ^ 2-2bc)] / 2
=[（a-b)^2+(a-c)^2+(b-c)^2]/2=0
Therefore, there must be: A-B = 0, a-c = 0, B-C = 0, that is, a = b = C, so △ ABC is an equilateral triangle

Given that the three sides ABC of a triangle satisfy the equation a to the cubic power + the cubic power of B + the cubic power of C = 3ABC, please determine the shape of the triangle Writing process

Equilateral triangle
a^3-abc+b^3-abc+c^3-abc=0
a(a^2-bc)+b(b^2-ac)+c(c^2-ac)=0
a. B and C cannot be equal to 0,
A ^ 2 = BC
b^2=ac
c^2=bc

Three sides a, B, C of triangle ABC satisfy the equation a square + C square-2ab-2bc + 2B square = 0, judge the shape of triangle ABC

(a-b)^2+(b-c)^2=0
a=b=c
Equilateral triangle

Given that the three sides a, B, C of the triangle ABC satisfy the equation A's square + C's Square - 2ab-2bc + 2B's Square = 0, try to judge the shape of triangle ABC

The square of a + the square of c-2ab-2bc + 2B = 0
(a-b）^2+(b-c)^2=0
a-b=0,b-c=0
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

We know that in the triangle ABC, the length of three sides a, B, C satisfies the equation a square-16b square-c square + 6ab + 10bc = 0, and prove that a + C = 2B

A square-16b square-c square + 6ab + 10bc = 0
=>(a+3b)^2-(c-5b)^2=0
=>A + 3B = c-5b or a + 3B = 5b-c
a+3b=c-5b
=>a-c+8b=0
=>C = a + 8b > A + B according to the relationship between the three sides of a triangle, it is impossible, so this relationship does not hold
So only a + 3B = 5b-c holds
=>a+c=2b

In the triangle ABC, we know that the triangle three sides a, B, C satisfy the equation (C ^ 2 / (a + b)) + (a ^ 2 / (B + C)) = B, and find the value of angle B How to solve the problem, what are the steps?

(C ^ 2 / (a + b)) + (a ^ 2 / (B + C)) = B, the denominator is removed to obtain the following results: C? (B + C) + a? (a + b) = B (a + b) (B + C) expansion: C? B + C ^ 3 + A ^ 3 + a? B = AB? + ABC + B ^ 3 + B

Given that a, B, C are the three sides of a triangle, a = 2n ^ 2 + 2n, B = 2n + 1, C = 2n ^ 2 + 2n + 1 (n is a natural number greater than 1), try to explain that △ ABC is a right triangle

a=2n^2+2n
b=2n+1
c=2n^2+2n+1
a^2=4n^4+8n^3+4n^2
b^2=4n^2+4n+1
c^2=4n^4+4n^2+1+8n^3+4n^2+4n=a^2+b^2
Such a triangle is a right triangle

Given that the three sides of the triangle ABC a, B, C satisfy the equations {2a-b = C, A-B + C = 6, a + B-C = 2}, try to explain that the triangle ABC is a right triangle

It is known that 2a-b = C ①, A-B + C = 6 ②, a + B-C = 2 ③,
② Add 3 to get 2A = 8, so a = 4,
① Add 3a-c of ③ = 2 + C, and substitute a = 4 to get C = 5,
So B = 3, because 3 ^ 2 + 4 ^ 2 = 5 ^ 2, the triangle ABC is a right triangle

Given that the triangles a, B, C satisfy the equation a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac = 0, try to judge the shape of △ ABC There should be an analysis process