It is known that A.B.C is the length of the three sides of the triangle ABC. If the square of B + 2Ab = the square of a + 2Ac is, judge what triangle ABC belongs to Try to explain that the square of a - the square of B + the square - 2Ac of C is greater than 0

It is known that A.B.C is the length of the three sides of the triangle ABC. If the square of B + 2Ab = the square of a + 2Ac is, judge what triangle ABC belongs to Try to explain that the square of a - the square of B + the square - 2Ac of C is greater than 0

It's obvious that you don't take special triangles at all
If the title is changed to: if the square of B + 2Ab = the square of C + 2Ac, then add the square of a to both sides
The square of (a + b) = the square of (a + C), B = C, isosceles triangle
For the supplement: the square of a - the square of B + the square of C - 2Ac = (A-C) the square of - B, and a-cps: sometimes there will be unexpected hints when you put some special numbers in

a. B. C is the three sides of triangle ABC, and b square + 2Ab = C square + 2Ac to judge the shape of triangle ABC

Therefore, it is an isosceles triangle

Given that ABC is the three sides of the triangle ABC, and the square of a is - 2BC = the square - 2Ac of B, try to judge the shape of the triangle ABC

The square of a-2bc = the square of b-2ac,
=>a^2-b^2=2bc-2ac
=>(a+b+2c)(a-b)=0
=>a-b=0
=>Triangle ABC is isosceles triangle

Let a, B and C be the lengths of the three sides of △ ABC, and B2 + 2Ab = C2 + 2Ac, then the shape of △ ABC is () A. Right triangle B. Isosceles triangle C. Isosceles right triangle D. Equilateral triangle

B2 + 2Ab = C2 + 2Ac can be changed to b2-c2 = 2ac-2ab,
（b+c）（b-c）=2a（c-b），
（b-c）（b+c+2a）=0，
We also know that B + C + 2A ≠ 0,
B = C,
So △ ABC is an isosceles triangle,
Therefore, B

a. B, C are the three sides of △ ABC. (1) when a square + 2Ab = C square + 2Ac, try to judge the shape of △ ABC. (2) prove that a square - b square + C square - 2Ac is less than 0

In triangle ABC
b^2+2ab=c^2+2ac
b^2-c^2+2ab-2ac=0
(b+c)(b-c)+2a(b-c)=0
(b-c)(2a+b+c)=0
Because a > 0, b > 0, C > 0 means 2A + B + C > 0
So B-C = 0, that is, B = C
The triangle is isosceles triangle
(2) It is known that a, B, C, are the three sides of △ ABC,
There are: a + b > C, A0, a-c-b

It is known that a, B, C are the three side lengths of △ ABC. When B2 + 2Ab = C2 + 2Ac, try to judge which kind of triangle △ ABC belongs to and explain the reason

∵b2+2ab=c2+2ac，
∴b2+2ab+a2=c2+2ac+a2，
∴（b+a）2=（c+a）2，
∵ a, B, C are the three sides of ᙽ ABC,
/ / A, B and C are all positive numbers,
∴b+a=c+a，
∴b=c，
The triangle is an isosceles triangle

It is known that a, B, C are the three side lengths of △ ABC. When B2 + 2Ab = C2 + 2Ac, try to judge which kind of triangle △ ABC belongs to and explain the reason

∵b2+2ab=c2+2ac，
∴b2+2ab+a2=c2+2ac+a2，
∴（b+a）2=（c+a）2，
∵ a, B, C are the three sides of ᙽ ABC,
/ / A, B and C are all positive numbers,
∴b+a=c+a，
∴b=c，
The triangle is an isosceles triangle

It is known that a, B, C are the three side lengths of △ ABC. When B2 + 2Ab = C2 + 2Ac, try to judge which kind of triangle △ ABC belongs to and explain the reason

∵b2+2ab=c2+2ac，
∴b2+2ab+a2=c2+2ac+a2，
∴（b+a）2=（c+a）2，
∵ a, B, C are the three sides of ᙽ ABC,
/ / A, B and C are all positive numbers,
∴b+a=c+a，
∴b=c，
The triangle is an isosceles triangle

It is known that ABC is the three sides of triangle ABC, and satisfies a-square-2ab = b-square-2ac, try to judge the shape of triangle ABC This is page 109 of Huanggang 100 people's education area

The triangle ABC is an equilateral triangle
It is proved that a ^ 2-2ab = B ^ 2-2ac
(a-b)²+2ac-2b²=0
(a-b)²+2(ac-b²)=0
Because a = b = C can satisfy the question
So the triangle ABC is an equilateral triangle

Given that ABC is the three sides of a triangle ABC and satisfies a ^ 2-2bc = B ^ 2-2ac, try to judge the shape of the triangle ABC

Δ ABC is an isosceles triangle for the following reasons:
∵a²-2bc=b²-2ac
∴a²-b²=2bc-2ac
∴(a-b)(a+b)=2c(b-a)
∴(a-b)(a+b+2c)=0
But a + B + 2C > 0
∴a-b=0
∴a=b
That is, △ ABC is an isosceles triangle