Q. 64.3( 6 Votes )

# If G be the centroid of a triangle ABC, prove that:

AB^{2} + BC^{2} + CA^{2} = 3 (GA^{2} + GB^{2} + GC^{2})

Answer :

We know that centroid of a triangle for (x_{1} , y_{1}), (x_{2} , y_{2}) and (x_{3} , y_{3}) is

G(x, y) = ( , )

We assume centroid of ∆ABC at origin.

For x=0 and y=0

= 0 and = 0

∴ = 0 and =0

Squaring on both sides, we get

x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + 2x_{1}x_{2} + 2x_{2}x_{3} + 2x_{3}x_{1} = 0 and y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + 2y_{1}y_{2} + 2y_{2}y_{3} + 2y_{3}y_{1} = 0 … (1)

AB^{2} + BC^{2} + CA^{2}

= [(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}] + [(x_{3} – x_{2})^{2} + (y_{3} – y_{2})^{2}] + [(x_{1} – x_{3})^{2} + (y_{1} – y_{3})^{2}]

= (x_{1}^{2} + x_{2}^{2} – 2x_{1}x_{2} + y_{1}^{2} + y_{2}^{2} – 2y_{1}y_{2})+(x_{2}^{2} + x_{3}^{2} – 2x_{2}x_{3} + y_{2}^{2} + y_{3}^{2} – 2y_{2}y_{3})+(x_{1}^{2} + x_{3}^{2} – 2x_{1}x_{3} + y_{1}^{2} + y_{3}^{2} - 2y_{1}y_{3})

= (2x_{1}^{2} + 2x_{2}^{2} + 2x_{3}^{2} – 2x_{1}x_{2} – 2x_{2}x_{3} – 2x_{1}x_{3}) + (2y_{1}^{2} + 2y_{2}^{2} + 2y_{3}^{2} – 2y_{1}y_{2} – 2y_{2}y_{3} – 2y_{1}y_{3})

= (3x_{1}^{2} + 3x_{2}^{2} + 3x_{3}^{2}) + (3y_{1}^{2} + 3y_{2}^{2} + 3y_{3}^{2}) …from 1

= 3(x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + 3(y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) … (2)

3(GA^{2} + GB^{2} + GC^{2})

= 3 [(x_{1} – 0)^{2} + (y_{1} – 0)^{2} + (x_{2} – 0)^{2} + (y_{2} – 0)^{2} + (x_{3} – 0)^{2} + (y_{3} – 0)^{2}]

= 3 (x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + x_{3}^{2} + y_{3}^{2})

= 3 (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + 3(y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) … (3)

From (2) and (3), we get

AB^{2} + BC^{2} + CA^{2} = 3(GA^{2} + GB^{2} + GC^{2})

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