# How do we prove that “the sum of squares of any of the two sides of a triangle equals to twice its square on half of the third side, along with the twice of its square on the median bisecting the third side”?

To prove this statement, “The sum of squares of any two sides of a triangle is equal to twice its square on half of the third side, along with twice of its square on the median bisecting the third side,” (this theorem is called Apollonius theorem) we can use the ...

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To prove this statement, “The sum of squares of any two sides of a triangle is equal to twice its square on half of the third side, along with twice of its square on the median bisecting the third side,” (this theorem is called Apollonius theorem) we can use the Pythagoras theorem.

First, let XYZ be a triangle in which XM is the median from the vertex 'X' to YZ.

We have to prove that XY^2+XZ^2= 2( XM ^2+(YZ/2)^2).

Proof: Let us draw XN perpendicular to YZ.

Since XM is the median, then YM= ZM = YZ/2

Also, as N is a point on YZ, then YN+NZ= YZ

From right triangle XYN, by Pythagoras theorem we have

XY^2= YN^2+XN^2, similarly in the right triangle XZN, we have

XZ^2= ZN^2+XN^2 and from the right triangle XNM: XM^2= XN^2+MN^2.

Adding first two equations we get XY^2+XZ^2= YN^2+XN^2+ZN^2+XN^2

= 2XN^2+YN^2+ZN^2, adding and subtracting 2MN^2

= 2XN^2 + 2MN^2 +YN^2+ZN^2-2MN^2

= 2(XN^2+MN^2)+ YN^2-MN^2+ZN^2-MN^2

= 2(XM^2)+ (YN+MN)(YN-MN)+ (ZN+MN)(ZN-MN)

= 2(XM)^2 + YM (YN -MN) + (ZN+MN) ZM

= 2(XM^ 2) + YZ/2* (YN-MN) + (ZN+MN)* YZ/2

= 2(XM)^2 + YZ/2 (YN-MN+ZN+MN)

= 2(XM)^2 + (YZ/2) (YN+ZN)

= 2(XM)^2 + (YZ /2)(YZ)

= 2(XM)^2 +(YZ)^2/2

= 2(XM)^2 + 2* (YZ^2)/4

= 2(XM)^2+ 2* (YZ/2)^2

= 2[(XM) ^2 + (YZ/2)^2]

Hence proved

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