If the three sides of the triangle ABC are ABC, the condition equation a squared + B squared + C squared = 6A + 8b + 10c-50. Try to judge the shape of the triangle

If the three sides of the triangle ABC are ABC, the condition equation a squared + B squared + C squared = 6A + 8b + 10c-50. Try to judge the shape of the triangle

If the three edges a, B and C of △ ABC satisfy the conditional equation a ^ 2 + B ^ 2 + C ^ 2 = 6A + 8b + 10c-50, try to judge the shape of △ ABC. Analysis: we should start from the conditional equation and find the relationship among the three edges of △ ABC. A ^ 2 + B ^ 2 + C ^ 2 = 6A + 8b + 10c-50, (a ^ 2-6a + 9) + (B2 -

The three sides a, B and C of △ ABC have the following relation formula: - C ^ 2 + A ^ 2 + 2ab-2bc = 0. Try to explain that this triangle is an isosceles triangle

The original formula =
a^2+b^2-b^2-c^2+2ab-2bc=0
(a+b)^2-(b+c)^2=0
a+b=b+c
So:
A=c
That is, the triangle is isosceles

Given that the three sides a, B, C of △ ABC satisfy the equation: a2-c2 + 2ab-2bc = 0, try to explain that △ ABC is isosceles triangle

∵a2-c2+2ab-2bc=0，
∴（a+c）（a-c）+2b（a-c）=0，
ν (A-C) (a + C + 2b) = 0, (2 points)
∵ a, B, C are the three sides of ∵ ABC,
A + C + 2B > 0, (3 points)
A-c = 0, there is a = C
Therefore, △ ABC is an isosceles triangle. (4 points)

The three sides a, B, C of the triangle ABC have the following relations: - C ^ + A ^ + 2ab-2bc = 0. Is a description; this triangle is an isosceles triangle

A, B and C of a triangle have the following relations: - the square of C + the square of a + 2ab-2bc = 0
Proof: this triangle is isosceles triangle
a^2+2ab=c^2+2bc
Add: B ^ 2
a^2+2ab+b^2=c^2+2bc+b^2
(a+b)^2=(c+b)^2
a+b=c+b
A=c
So this triangle is an isosceles triangle

We know that the length of the three sides of the inequality △ ABC is an integer a, B, C, and satisfies the requirements of a 2 + B 2 - 4a-6b + 13 = 0

∵a²+b²-4a-6b+13,
=a²-4a+4+b²-6b+9,
=（a-2）²+（b-3）²=0,
∴a-2=0,b-3=0,
A = 2, B = 3,
∵3-2=1,3+2=5,
∴1＜c＜5,
The length of the three sides of ABC is positive integers a, B, C,
∴c=4．

It is known that the three side lengths of △ ABC are a, B, C respectively, and a, B, C satisfy the equation: a? + B? + C? + 2 (AB BC AC) = 0 To judge the shape of △ ABC, the variation a 2 + B 2 + C 2 - AB BC AC = 0

a²+b²+c²+2(ab-bc-ac）=0
a²+b²+c²+2ab-2bc-2ac=0
(a²+b²+2ab)+(-2bc-2ac)+c²=0
(a+b)²-2(a+b)c+c²=0
(a+b-c)²=0
a+b-c=0,
Because a, B and C are the three sides of a triangle, the sum of the two sides cannot be equal to the third side
a²+b²+c²-ab-bc-ac=0
2a²+2b²+2c²-2ab-2bc-2ac=0
(a²+b²-2ab)+(a²+c²-2ac)+(b²+c²-2bc)=0
(a-b )²+(a-c)²+(b-c)²=0
So, A-B = 0, a-c = 0, B-C = 0
So a = b = C
Equilateral triangle

It is known that the three sides of △ ABC are a, B and C respectively, and a, B and C satisfy the equation a 2 - C 2 + AB BC = 0, so as to determine the shape of the triangle

a²-c²+ab-bc=0
(a+c)(a-c)+b(a-c)=0
(a-c)(a+c+b)=0
a. B, C are the sides of the triangle, which are always positive, and a + C + B are always positive. If the equation holds, only a-c = 0
A = C, a triangle is an isosceles triangle with B as the base and a and C as the waist

It is known that in △ ABC, the length of three sides a, B and C satisfies the equation a 2 - 16 B 2 - C 2 + 6 AB + 10 BC = 0. It is proved that a + C = 2 b

a²-16b²-c²+6ab+10bc=0
a²+6ab+9b²-25b²-c²+10bc=0
（a²+6ab+9b²）-（25b²+c²-10bc）=0
（a+3b）²-(5b-c)²=0,
（a+3b）²=(5b-c)²,
So a + 3B = 5b-c, or a + 3B = - (5b-c), (omit)
That is, a + C = 2B

It is known that the three side lengths of △ ABC are a, B, C respectively, and a, B, C satisfy the equation: a? + B? + C? = AB + AC + BC Try to judge the shape of the triangle

Δ ABC is an equilateral triangle
∵a2+b2+c2-ab-bc-ac=0,
∴2a2+2b2+2c2-2ab-2bc-2ac=0,
∴a2-2ab+b2+b2-2bc+c2+a2-2ac+c2=0,
That is (a-b) 2 + (B-C) 2 + (C-A) 2 = 0,
∴a-b=0,b-c=0,c-a=0,
∴a=b=c,
The △ ABC is an equilateral triangle

It is known that a, B, C are the three sides of △ ABC and satisfy the relation a  2 + C  = 2Ab + 2bc-2b  to judge the shape of △ ABC

∵a²+c²=2ab+2bc-2b²
∴（a-b)²+（c-b)²=0
∴a=b=c
The equilateral triangle