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.
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 = 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.
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).
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.
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.