NOTE- this question is not related to the topic which I have selected.
M is any point in the plane of triangle ABC. prove that AM² + BM² + CM² = AG² + BG² + CG² + 3 MG² , where G is centroid of triangle ABC.

Open in App

Solution

Let the coordinates of A, B and C be (x₁.y₁), (x₂.y₂) and (x₃.y₃)
Then coordinates of centroid G will be ((x₁ + x₂ + x_{3)/3, (}y₁ + y₂ + y_{3)/3 )}
Let us the assume that centroid lies on the origin
⇒ x₁ + x₂ + x₃ = 0 and y₁ + y₂ + y₃ = 0
Let the coordinates of point M be (x, y)
We have to prove AM² + BM² + CM² = GA² + GB² + GC² + 3GM²
Now, L.H.S. = MA² + MB² + MC²
(x – x₁)² + (y – y₁)² + (x – x₂)² + (y – y₂)² + (x – x₃)² + (y – y₃)²
= 3x² + 3y² + x₁² + x₂² + x₃² + y₁² + y₂² + y₃² – 2x (x₁ + x₂ + x₃) – 2y(y₁ + y₂ + y₃)
= 3x² + 3y² + x₁² + x₂² + x₃² + y₁² + y₂² + y₃²
R.H.S. = GA² + GB² + GC² + 3GM²

= (x₁ – 0)² + (y₁ – 0)² + (x₂ – 0)² + (y₂ – 0)² + (x₃ – 0)² + (y₃ – 0)² + 3[(x – 0)² + (y – 0)²]
= x₁² + y₁² + x₂² + y₂² + x₃² + y₃² + 3(x² + y²)
= 3x² + 3y² + x₁² + x₂² + x₃² + y₁² + y₂² + y₃²
= L.H.S.

Suggest Corrections

Similar questions

G

A B C

\frac{A G^{2} + B G^{2} + C G^{2}}{A B^{2} + B C^{2} + C A^{2}} = \frac{1}{3}

G

△ A B C

\frac{A B^{2} + B C^{2} + C A^{2}}{A G^{2} + B G^{2} + C G^{2}}

A B C

B

M

B C

A M^{2} + B C^{2} = A C^{2} + B M^{2}

P A^{2} + P B^{2} + P C^{2} = G A^{2} + G B^{2} + G C^{2} + 3 G P^{2}

A M^{2} + B C^{2} = A D^{2} + B M^{2}

Join BYJU'S Learning Program