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3 (square of a + square of B + square of C) = (a + B + C) = square of (a + B + C) + 2Ab + 2BC + 2ca 2 (a 2 + B 2 + C? - 2Ab + 2BC + 2ca = 02 (a? + B? + C?) - 2Ab + 2BC + 2ca = (a-b) 2 + (B-C) 2 + (C-A) 2) = 0 because

It is known that the three sides of the triangle ABC are ABC respectively, and a, B and C satisfy the equation 3 (the square of a + the square of B + the square of C) = (a + B + C)

∵ 3(a²+b²+c²)=(a+b+c)² ,∴ 3a²+3b²+3c²=a²+b²+c²+2ab+2bc+2ac,∴ 2a²+2b²+2c²-2ab-2bc-2ac=0 ,(a²-2ab+b²)+(b²-2bc+c²)+...

In the triangle ABC, we know that the three sides a, B, C satisfy the square of B + the square of a - the square of C = AB, and find the angle C

cosC=(a²+b²-c²)/2ab=ab/2ab=1/2,C=60°

ABC is the three side length of triangle ABC, a = 2n square + 2n, B = 2n + 1, C = 2n square + 2n + 1 (n is a natural number), judge whether the triangle ABC is a right angle Given that A.B.C is the three side length of triangle ABC, a = 2n square + 2n, B = 2n + 1, C = 2n square + 2n + 1 (n is natural number), try to judge whether triangle ABC is a right triangle and explain the reasons

a^2+b^2=4n^4+8n^3+8n^2+4n+1
c^2=4n^4+8n^3+8n^2+4n+1
a^2+b^2=c^2
Δ ABC is a right triangle

It is known that a, B and C are the three sides of a triangle, a = 2n 2 + 2n, B = 2n + 1, C = 2n 2 + 2n + 1 (n is a natural number greater than 1), we try to explain that △ ABC is a right triangle

Because n is a natural number greater than 1, C is the longest side
∵a2+b2=4n4+8n3+8n2+4n+1，
c2=4n4+8n3+8n2+4n+1，
∴a2+b2=c2，
The △ ABC is a right triangle

In △ ABC, the length of three sides is a continuous natural number, and the maximum angle is 2 times of the minimum angle

Trilateral X-1, x, x + 1
The two angles are a and 2A
Then 2a to x + 1, a to X-1
sin2a=2sinacosa
By sine theorem
(x-1)/sina=(x+1)/sin2a=(x+1)/2sinacosa
So X-1 = (x + 1) / 2cosa
cosa=(x+1)/2(x-1)
From cosine theorem
cosa=[(x+1)^2+x^2-(x-1)^2]/2x(x+1)
[(x+1)^2+x^2-(x-1)^2]/2x(x+1)=(x+1)/2(x-1)
2(x-1)(x^2+4x)=2x(x+1)^2
2x(x-1)(x+4)=2x(x+1)^2
x^2+3x-4=x^2+2x+1
X=5
So the three sides are four, five, six

In △ ABC, a = m2-n2, B = 2Mn, C = M2 + N2, where m and N are positive integers; and M > N, try to judge whether △ ABC is a right triangle?

∵a=m2-n2，b=2mn，c=m2+n2，
∴a2+b2=（m2-n2）2+4m2n2=m4+n4-2m2n2+4m2n2=m4+n4+2m2n2=（m2+n2）2=c2．
The △ ABC is a right triangle

In △ ABC, a = N2, B = n2-1 / 2, C = N2 + 1 / 2, where n is a positive odd number. It is proved that this triangle is a right triangle

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As shown in the figure, the side length of the small square is 1

It is proved that: ac2 = 12 + 22 = 5, BC2 = 12 + 22 = 5, AB2 = 12 + 32 = 10,
∴AC2+BC2=AB2=10，AC=BC=
5，
The △ ABC is an isosceles right triangle

Are triangles of 2n2 + 2n, 2n + 1, 2n2 + 2n + 1 (n > 0) right angles? Why?

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