Knowing a b c, they have the following relation a b c 0, b 0, and a b c , which is the symbol for ju

a|+|b|=|c|,|a|-b=|c|,, so the absolute value of c is the largest, and a+b+c 0 so c is greater than zero.

And |b|Less than |c|,|a|+|b|=|c|, so |a|Less than |c|, so a is greater than zero.

then A and C have the same sign.

Hope, thank you

The solution can be obtained from the meaning of the question: a+b+c 0, b 0

So a+b>-c

And because |a|+|b|=|c|, so we can launch a,b different name a>0 so |a|+|b|=a-b=c or -c

When a-b=-c, we get a+c=b<0 which contradicts what we know a+b+c 0.

So a-b=c>0

So a>0, c>0

Solution: c|-|a|=|b|

and 0>b>-a-c

Therefore, there is: a+c|>|b|>0

Therefore|a+c|>|c|-|a| （

Both sides are squared and organized.

ac>0

Therefore, AC has the same name.

and from a+b+c>0, b<0

Therefore, it is known that a>0, c>0

Because iai+ibi=ici, when b<0 and c<0, the sum of the three must be less than zero, which is inconsistent with the original condition a+b+c>0, so c>0

In the same way, it can be proved that if a<0, then the sum of the three is equal to zero, which is inconsistent with the original condition a+b+c>0, so a>0

According to Yu Shensen's string theorem: Kuanshi mu a 2 = b 2 + c 2-2bc * cosa is obtained by the title (a + c) (a-c) = b (b + c).

b^2+c^2-a^2)=-bc;

From the cosine theorem, cosa=(b 2+c 2-a 2) 2bc=-1 2; The approved manuscript is cosa=-1 2; So a = 120 degrees.

Because a 0, b 0, so a+b 0 and because b 0, c 0, so c-b 0, b-c 0 is positive according to the multiplication of two numbers with the same sign, and the different sign is negative (a+b)(c-b) 0, (a+b)(b-c) 0, so (a+b)(c-b) (a+b)(b-c).

The answer is b

Note that there is no cornering principle. Drawing a line from an obtuse angle in an obtuse triangle and constructing an isosceles triangle with a short side as one side proves that the edge is incorrect. Because a large obtuse triangle and a small obtuse triangle are two sides and one angle equal.

ab=ac ad=ad d= d, but bad and cad are not congruent in the same way, and there is no corner edge principle to prove congruence.

Solution: A, correct, in line with SAS judgment;

b, incorrect, because the sides bc and b c are not on the side of a and a, so two triangles cannot be congruent;

c. Correct, in line with AAS judgment;

d, correct, in accordance with the ASA judgment;

abc 0, and b grip 0, ac bright call 0, a c, segment key eggplant.

a＞0 c＜0．

Sine theorem. b／sinb＝c／sinc

b＝csinb／sinc

Gen3 Gen3 2) (Gen6 Gen2) 4

6 (root 6, ten root 2).

3 (root 6 a 2) 2

3 roots 6 3 roots 2) 2

A=a-b, b=c-b then the original formula becomes:

a-b+c-b)^2-4(a-b)(c-b)=(a+b)^2-4ab=(a-b)^2=0

So a=b, i.e. a-b=c-b

So a=c