Given that a, B and C are the lengths of the three sides of △ ABC and satisfy A2 + 2B2 + c2-2b (a + C) = 0, then the shape of the triangle is______ ．

Given that a, B and C are the lengths of the three sides of △ ABC and satisfy A2 + 2B2 + c2-2b (a + C) = 0, then the shape of the triangle is______ ．

From the known condition A2 + 2B2 + c2-2b (a + C) = 0, it is concluded that, (a-b) 2 + (B-C) 2 = 0  A-B = 0, B-C = 0, that is, a = B, B = C  a = b = C, so the answer is equilateral triangle

The lengths of three sides of △ ABC are a, B and C respectively, if B & # 178; + C & # 178; - BC = a (B + C-A)

b²+c²-bc=a(b+c-a)
b²+c²-bc=ab+ac-a²
2a²+2b²+2c²-2ab-2bc-2ca=0
(a-b)²+(b-c)²+(c-a)²=0
So A-B = B-C = C-A = 0
That is, a = b = C
ABC is an equilateral triangle

In Δ ABC, a, B and C are the lengths of three sides of a, B and C respectively. It is known that a, B and C satisfy B & # 178; = AC and a & # 178; - C & # 178; = AC BC,
Finding the size of a and the value of bsinb / C

b²=ac ①,a²－c²=ac－bc ②
① - 2, B & # 178; - A & # 178; + C & # 178; = BC
And B & # 178; - A & # 178; + C & # 178; = 2BC Cosa # cosa = 1 / 2 # a = 1 / 3 π
∵b²=ac,∴sinB×b/c=sinB×a／b=sinB×sinA／sinB=sinA=√3／2
If you can,