It is known that a, B and C are the three sides of △ ABC, and the value of a, B and C is such that the value of fraction (AB AC + the square BC of C) A-B is 0. Judge the shape of the triangle and explain the reason

It is known that a, B and C are the three sides of △ ABC, and the value of a, B and C is such that the value of fraction (AB AC + the square BC of C) A-B is 0. Judge the shape of the triangle and explain the reason

(ab-ac+c^2-bc)/(a-b)=0
The denominator is not zero, A-B ≠ 0, a ≠ B
ab-ac+c^2-bc=0
a(b-c) -c(b-c) =0
(a-c)(b-c)=0
A = C, or B = C, isosceles triangle

1. Given ABC ≠ 0, find the value of a / | a | + B / | B | + C / | C |
2 given ABC ≠ 0 and a + B + C = 0, find the value of a / | a | + B / | B | + C / | C |
How do you understand that?

∵abc≠0
| A / | a | = 1 (a > 0) or - 1 (A0) or - 1 (B0) or - 1 (c

If | a | = 5.5 | B | = 2.5 | C | = 5, find the value of ABC

A:
|a|=5.5
|b|=2.5
|c|=5
The result of multiplication of the three formulas is as follows
|abc|=5.5×2.5×5=68.75
So:
ABC = 68.75 or ABC = - 68.75

Given a + B / a = a + C / b = B + C / A, find the value of ABC / (a + b) (a + C) (B + C)

1/8