Question

if the sides of triangle ABC are 5,7,8 units then AG^2+BG^2+CG^2=

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

i) By Apollonious theorem, "In a triangle ABC, AB² + AC² = 2(AD² + BD²),

where AD is the median to the side BC.

ii) By property of the Centroid of a triangle, it divides eah median in the ration 2:1

So AG = (2/3)AD; hence, AD = 3AG/2

Also BD = (1/2)(BC), since D is the midpoint of BC, AD being the median,

iii) Substituting the above in (i),

AB² + AC² = 2(9AG²/4 + BC²/4) =

Multiplying by 2, 2AB² + 2AC² = 9AG² + BC²

==> 2AB² + 2AC² - BC² = 9AG²

Similarly, 2BC² + 2AB² - AC² = 9BG²

and 2BC² + 2AC² = 9CG²

iv) Adding all the 3 above,

3(AB² + BC² + CA²) = 9(AG² + BG² + CG²)

==> AB² + BC² + CA² = 3(AG² + BG² + CG²) [Proved]