If the three sides of the triangle ABC are a, B, C (1) the second power of a-C + 2ab-2bc = 0, it is proved that the triangle ABC is an isosceles triangle 2. If the second power of a + the second power of B + the second power of C - AB BC AC = 0, judge the shape of triangle ABC

If the three sides of the triangle ABC are a, B, C (1) the second power of a-C + 2ab-2bc = 0, it is proved that the triangle ABC is an isosceles triangle 2. If the second power of a + the second power of B + the second power of C - AB BC AC = 0, judge the shape of triangle ABC

The second power of a - the second power of C + 2ab-2bc = (a + C) (A-C) + 2B (A-C) = (A-C) (a + 2B + C) = 0
Because (a + 2B + C) is not equal to 0, a-c = 0, a = C triangle ABC is an isosceles triangle
a²+b²+c²-ab-ac-bc=0
2a²+2b²+2c²-2ab-2ac-2bc=0
a²-2ab+b²+a²-2ac+c²+c²-2bc+b²=0
（a-b）²+（a-c）²+（c-b）²=0
a=b=c
Equilateral triangle

In △ ABC, the opposite sides of angles a, B and C are a, B, C, A2 + C2 − B2 = 1, respectively 2ac． (I) find sin2a + C 2 + cos2b; (II) if B = 2, find the maximum of △ ABC area

(I) according to cosine theorem: CoSb = 14sin2a + C2 + cos2b = sin2 (π 2 − B2) + 2cos2b − 1 = cos2b2 + 2cos2b − 1 = 1 + cosb2 + 2cos2b − 1 = − 14 (II) SINB = 154 ∵ B = 2, A2 + C2 − B2 = 12ac ﹤ A2 + C2 = 12ac + B2 = 12ac + 4 ≥ 2Ac, thus AC ≤ 83

Given that A.B.C is the three sides of the triangle ABC and satisfies the relation | 2a-b-1 | = - (A-2) square, C is an even number

If the absolute value term and the square term are not negative and are opposite to each other, then they are both equal to 0
2a-b-1=0
a-2=0
A = 2 b = 3 is obtained
The sum of the two sides of the triangle > the third side, the difference between the two sides

It is known that the three sides of the triangle ABC are a, B, C and satisfy the judgment of the relation a square + b square + C square + 50 = 6A + 8A + 10C It is known that the three sides of the triangle ABC are a, B, C and satisfy the relation a square + b square + C square + 50 = 6A + 8A + 10C Judging the shape of triangle ABC

A square + b square + C square + 50 = 6A + 8b + 10Ca square + b square + C square + 50-6a-8b-10c = 0 (A-3) 2 + (B-4) 2 + (C-5) 2 = 0a-3 = 0b-4 = 0c-5 = 0A = 3, B = 4, C = 53? + 4? = 5? A? + B? = C? So, the triangle ABC is a right triangle

If the triangle ABC satisfies the relation (a-b) square + (a-b) C = 0, then the triangle must be a () triangle?

Because 0 = (a-b) ^ 2 + (a-b) C = (a-b) (a-b + C), because the sum of the two sides of the triangle is greater than the third side, there must be a = B, which is an isosceles triangle

There is a relationship between ABC and ABC - C square + a square + 2ab-2bc = 0. Try to prove that this triangle is isosceles triangle

∵ - C square + a square + 2ab-2bc = 0
∴（a+c)(a-c)+2b(a-c)=0
(a-c)(a+c+2b)=0
∵a+c+2b≠0
∴a-c=0
∴a=c
That is, the triangle is an isosceles triangle

Given that the three sides a, B, C of △ ABC satisfy the equation: a2-c2 + 2ab-2bc = 0, try to explain that △ ABC is isosceles triangle

∵a2-c2+2ab-2bc=0，
∴（a+c）（a-c）+2b（a-c）=0，
ν (A-C) (a + C + 2b) = 0, (2 points)
∵ a, B, C are the three sides of ∵ ABC,
A + C + 2B > 0, (3 points)
A-c = 0, there is a = C
Therefore, △ ABC is an isosceles triangle. (4 points)

A.b.c. has the following relation: - C2 + A2 + 2ab-2ac = 0 -The square is C2

a²-c²+2ab-2ac=0
(a+b)²-(a+c)²-b²+a²=0
(a+b)²+a²=(a+c)²+b²
①:
a+b=a+c
∴b=c
② : a + B = B a = 0 (round off)
Ψ is an isosceles triangle

It is known that a, B, C is the three side length of triangle ABC, and the square of (B-C) = (- 2a-b) (C-B). It is shown that triangle ABC is an isosceles triangle

(b-c)²=（-2a-b)(c-b)
b²-2bc+c²=-2ac+2ab-bc+b²
-bc+c²+2ac-2ab=0
c(c-b)+2a(c-b)=0
(c+2a)(c-b)=0
Because a, B and C are all > 0, C-B = 0, b-c

Given that a, B and C are the three sides of the triangle ABC respectively, try to judge the sign of the algebraic formula (a 2 + B? - C 2) - 4A? B Explain why

（a²+b²-c²）²-4a²b²
=(a²+b²-c²+2ab)(a²+b²-c²-2ab)
=[(a+b)²-c²][(a-b)²-c²]
=(a+b-c)(a+b+c)(a-b+c)(a-b-c)
The sum of the two sides of the triangle is greater than the third
a,b,c>0 ==>a+b+c>0
a+b>c ==>a+b-c>0
a+c>b ==> a-b+c>0
a a-b-c