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

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

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

1. If the right angle side of a right triangle is 8,15, then the length of the hypotenuse side is -- and the length of the hypotenuse is -- 2. If the length of three sides a, B, C of triangle ABC satisfies the relation (a + 2b-60) 2 + | B-18 | + | c-30 | = 0, then the three sides of triangle ABC are respectively a=___ ,b=____ ,c=___ The shape of triangle ABC is___ . 3. A 25 decimeter long ladder is inclined to a vertical wall. This is the distance between the ladder foot and the bottom of the wall. If the top of the ladder falls 4 decimeters along the wall, the foot will slide____ . 4. As shown in the figure, a flagpole is broken at 9m above the ground, and the top of the flagpole is 12m away from the bottom of the flagpole, which is higher before the flagpole is broken___ M 5. Xiaoming's home is 1000m away from the school, 800m away from the bookstore and 600m away from the school. Question: is the triangle composed of Xiaoming's home, school and bookstore as its apex? Why

（2）∵（a+2b-60）²+|b-18|+|c-30|＝0
∴a+2b-60=0,b-18=0,c-30=0
∴a+2b=60,①
b=18,②
c=30
Replace ② into ①: a + 2B = 60
a+2×18=60
a+36=60
∴ a=42
∴ A=42 B=18 C=30
(1) The solution ∵ is a right triangle
The length of right angle side is 8,15
The length of the hypotenuse is 17 (the inverse theorem of Pythagorean theorem)
……

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

It is known that a, B, C are the three sides of △ ABC, and a  2 + 2Ab = C  2 + 2BC. It is proved that △ ABC is an isosceles triangle

prove:
a²+2ab=c²+2bc
a²+2ab+b²=c²+2bc+b²
(a+b) ²=(b+c) ²
(a+b) ²-(b+c) ²=0
(a+b+b+c)(a-b-b-c)=0
(a+2b+c)(a-c)=0
If a + 2B + C > 0, then a-c = 0
Then a = C
Then △ ABC is an isosceles triangle

It is proved that ABC is an isosceles triangle if the trilateral a B C of △ ABC and - C 2 ＋ a ＋ 2Ab minus 2BC = 0

﹣c²＋a²＋2ab－2bc＝0
﹙a²－c²﹚＋2b﹙a－c﹚＝0
﹙a＋c﹚﹙a－c﹚＋2b﹙a－v﹚=0
﹙a－c﹚﹙a＋c＋2b﹚=0
a－c=0
A=c
ABC is an isosceles triangle

Given that the three sides of △ ABC are a, B, C, and satisfy the equation (a-b) 2 + (B-C) 2 + (C-A) 2 = 0, tell the shape of the triangle,

∵（a-b）²+（b-c）²+（c-a）²=0
∴a-b=b-c=c-a=0
∴a=b=c
The △ is an equilateral triangle

It is known that the length of the three sides of △ ABC is a B C, which satisfies the equation (a-b) 2 + (B-C) 2 + (C-A) 2 = 0, Try to guess the shape of △ ABC and explain the reason

It can be concluded from the meaning of the title
a=b=c
So the triangle is equilateral

When the square of B + 2Ab = the square of C + 2Ac, the shape of the triangle ABC is judged The sign of judging the square of A-B and the square-2ac of C

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

a. If a ^ 2 + 2Ab = C ^ 2 + 2BC, try to judge the shape of △ ABC. (2) prove a ^ 2-B ^ 2 + C ^ 2-2ac < 0

1. Therefore, a-c = 0A = C isosceles triangle 2, a-c + b-2ac = (A-C) + 2B (A-C) = 0 (A-C) (a + C + 2b) = 0A + C + 2B > 0, so a-c = 0A = C isosceles triangle 2, a-c + B ^ - 2Ac = (A-C) 2 ^ = (A-C + b) (A-C-B) the sum of the two sides of the triangle is greater than the third side, so a-c + b > 0a-c-b