If G be the centroid of a triangle ABC, prove that, AB2+BC2+CA2=3(GA2+GB2+GC2) - Maths - Coordinate Geometry

Question

If G be the centroid of a triangle ABC, prove that,
AB2+BC2+CA2=3(GA2+GB2+GC2)

Lalit Mehra · Accepted Answer

Hi!
Here is the answer to your question
Let A (x1, y1),
B(x2, y2) and
C(x3, y3), be the vertices of
âˆ†ABC
Without the loss of Generality, assume the centroid of the
ΔABC to be at the origin, i e G = (0, 0)
<img height="83" width="331" src=
"https://s3mn%20mnimgs%20com/img/shared/userimages/mn_images/image/a2(19)%20png"
alt="">
⇒ x1 + x2 +
x3 = 0 and y1 +
y2 + y3 = 0
Squaring on both sides, we get
x12 +
x22 + x32
+ 2x1x2 +
2x2x3 +
2x3x1 = 0 and
y12 + y22
+ y32 +
2y1y2 +
2y2y3 +
2y3y1 = 0 … (1)
AB2 + BC2 + CA2
= [(x2 –
x1)2 + (y2 –
y1)2] + [(x3 –
x2)2 + (y3 –
y2)2] + [(x1 –
x3)2 + (y1 –
y3)2]
= [(x12 +
x22 –
2x1x2 +
y12 + y22
– 2y1y2) +
(x22 +
x32 –
2x2x3 +
y22 + y32
– 2y2y3) +
(x12 +
x32 –
2x1x3 +
y12 + y32
– 2y1y3)
= (2x12 +
2x22 +
2x32 –
2x1x2 –
2x2x3 –
2x1x3) +
(2y12 +
2y22 +
2y32 –
2y1y2 –
2y2y3 –
2y1y3)
= (3x12 +
3x22 +
3x32) +
(3y12 +
3y22 +
3y32) (From (1))
= 3(x12 +
x22 +
x32) +
3(y12 +
y22 +
y32) … (2)
3(GA2 + GB2 + GC2)
= 3 [(x1 – 0)2 +
(y1 – 0)2 +
(x2 – 0)2 +
(y2 – 0)2 +
(x3 – 0)2 +
(y3 – 0)2]
= 3 (x12 +
y12 + x22
+ y22 +
x32 +
y32)
= 3 (x12 +
x22 +
x32) +
3(y12 +
y22 +
y32) … (3)
From (2) and (3), we get
AB2 + BC2 + CA2 =
3(GA2 + GB2 + GC2)
Cheers!