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If G be the centroid of a triangle ABC, prove that, AB2+BC2+CA2=3(GA2+GB2+GC2). Please explain with detailed solution.

If G be the centroid of a triangle ABC, prove that, AB2+BC2+CA2=3(GA2+GB2+GC2). Please explain with detailed solution.

Grade:11

2 Answers

Arun
25758 Points
4 years ago
 
Let A (x1, y1), B(x2, y2) and C(x3, y3), be the vertices of ∆ABC.
 
Without the loss of Generality, assume the centroid of the ΔABC to be at the origin, i.e. G = (0, 0).
 
Centroid of triangle ABC = [x1 +x2+x3/3, y1 +y2+y3/3]
 
Hence
x1 +x2+x3 = 0  and  y1+y2+y3 = 0
 
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)
 
 
Regards
Arun (askIITians forum expert)
Kushagra Madhukar
askIITians Faculty 629 Points
2 years ago
Dear student,
Please find the solution to your problem below.
 
Let A (x1, y1), B(x2, y2) and C(x3, y3), be the vertices of ∆ABC.
 
Without the loss of Generality, assume the centroid of the ΔABC to be at the origin, i.e. G = (0, 0).
 
Centroid of triangle ABC = [x1 +x2+x3/3, y1 +y2+y3/3]
 
Hence
x1 +x2+x3 = 0  and  y1+y2+y3 = 0
 
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)
 
Thanks and regards,
Kushagra

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