# Prove that the triangle is equilateral if ,2a^2+b^2+c^2=2a(b+c), when a,b,c are the lenghts of the sides of a triangle.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If a,b,c, are the lengths of the sides of an equilateral triangle, that means that:

a = b = c

So, one way to solve the problem is to substitute b and c by a, in the given relation:

2a^2 + a^2 + a^2 = 2a(a+a)

We'll combine like terms and we'll get:

4a^2 = 2a*2a

4a^2 = 4a^2 q.e.d.

Another method of solving the proble is to remove the brackets from the right side:

2a^2+b^2+c^2=2a(b+c)

2a^2+b^2+c^2 = 2ab + 2ac

We'll subtract both sides 2ab + 2ac:

2a^2+b^2+c^2 - 2ab - 2ac = 0

We'll write 2a^2 = a^2 + a^2

We'll combine the terms in such way to complete the squares:

(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) = 0

(a-b)^2 + (a-c)^2 = 0

We'll impose the constraint of an equilateral triangle, that:

a = b = c

and we'll substitute b by a and c by a:

(a-a)^2 + (a-a)^2 = 0

0 + 0 = 0

0 = 0 q.e.d.

So, the relation holds if and only if the triangle is equilateral.

neela | High School Teacher | (Level 3) Valedictorian

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The sides of a triangle are give to be a, b, c.

To prove that  the triangle is eqilateral if 2a^2+b^2+c^2 = 2a(b+c).

Proof.

When the triangle is equilateral, a = b =c..

So it is sufficient if we replace b by a and c by a and if this verifies the 2a^2+b^2+c^2 = 2a(b+c).

So the LHS = 2a^2+b^2+c^2 = 2a^2+a^2+a^2 = 4a^2.

RHS : 2a(b+c) = 2a(a+a)  = 4a^2.

Therefore  2a^2+b^2+c^2 = 2a(b+c), when the sides a, b,c are  the sides of an equilateral triangle.

william1941 | College Teacher | (Level 3) Valedictorian

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If we can prove that the expression 2a^2 + b^2 + c^2 is equal to 2a(b+c) for a= b= c, we can prove that the triangle is an equilateral triangle.

Now 2a^2 + b^2 + c^2 -  2a(b+c)

=> 2a^2 + b^2 + c^2 - 2ab - 2ac

Now impose the condition that a= b= c,

we get : 2a^2 + a^2 + a^2 - 2a^2 - 2a^2 = 0

=> 2a^2 + a^2 + a^2 = 2a^2 + 2a^2

=> 2a^2 + b^2 + c^2 = 2ab + 2ac

=> 2a^2 + b^2 + c^2 = 2a(b+c)

So the required result follows.