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If A.B.C is the three sides of the triangle ABC, and the square of a + the square of B = AB + BC + Ca, then the triangle ABC is equilateral
A ^ 2 + B ^ 2 + C ^ 2 = AB + BC + AC both sides are the same multiplication 2: 2A ^ 2 + 2B ^ 2 + 2C ^ 2 = 2Ab + 2BC + 2Ac the right side of the equal sign is moved to the left side, and the result is as follows: (a ^ 2 + B ^ 2-2ab) + (C ^ 2 + A ^ 2-2ac) + (b ^ 2 + C ^ 2-2bc) = 0 so: (a-b) ^ 2 + (A-C) ^ 2 + (B-C) ^ 2 = 0 becuase: the complete equality of a number
It is known that ABC is the three sides of the triangle ABC, and the square of a + the square of bc-ac-b = 0
a^2+bc-ac-b^2=(a+b)(a-b)+c(b-a)=(a-b)(a+b-c)=0
So A-B = 0, a = B or a + B-C = 0
Because the sum of any two sides of the triangle is greater than the third side, a + B-C = 0 does not exist
Therefore, among the three sides of a triangle, a = B, the triangle is an isosceles triangle
Given that the three sides of the triangle ABC are a, B, C and satisfy the relation a square + b square + C square = AB + BC = Ca, try to judge the triangle ABC and explain the reason
And satisfy the relation a square + b square + C square = AB + BC = ca
It should be "and satisfy the relation a square + b square + C square = AB + BC + Ca"
Triangle ABC is an equilateral triangle
∵a²+b²+c²=ab+bc+ca,
∴2a²+2b²+2c²=2ab+2bc+2ca,
2a²+2b²+2c²-2ab-2bc-2ca=0
（a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)=0
(a-b)²+(a-c)²+(b-c)²=0
∴(a-b)²=0,(a-c)²=0,(b-c)²=0
∴a-b=0,a-c=0,b-c=0
∴a=b=c,
That is, △ ABC is an equilateral triangle
It is known that the three sides of the triangle ABC are A.B.C, and the square of a + the square of B + the square of C = AB + BC + ca It is proved that the triangle is equilateral
A 2 + B 2 + C 2 = AB + BC + CAA 2 + B 2 + C 2 - ab - BC - AC = 0, the two sides are multiplied by 22a 2 + 2B 2 + 2C 2 a B - 2 ab - 2 BC - 2 AC = 0 (a 2 a B + B 2) + B 2 B C + C 2) + (C 2 - 2 AC + a 2) = 0 (a - b) 2 + (B - C)
If the three sides of the triangle ABC are A.B.C and meet the condition that the square of a + the square of B + the square of C = AB + BC + Ca, why is the triangle% d Emergency amount... But it's better to have a process
On both sides of the equation, we can get (a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0, so a = b = C
Note: A ^ 2 is the square of A
If a ^ 2-6a + B ^ 2-8b + 25 = 0, judge the shape of triangle ABC Question 2 The length of both sides of the isosceles triangle ABC is two real roots of the equation x ^ 2-mx + 3 = 0 about X. it is known that the length of one side of the isosceles triangle ABC is 3, and its perimeter is calculated Although not many points, full of sincerity sorry~c=5
a a a a a ? a a a a a a ? a ? a ? a ? a ? a ? a ? a ? a
If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle
According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle
If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle
According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle
If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle
0
0
According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle
RELATED INFORMATIONS
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- 13. 0
- 14. 0
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- 18. 0
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