Question

if ad,be and CF are medians of the triangle abc,then prove that 3(ab^2+bc^2+ca^2)=4(ad^2+be^2+cf^2)

Answer 1

Here we use the relation between the lengths of the sides and that of the sides. Therefore we get:

2*CF^2= AC^2 + BC^2 - (1/2)*AB^2

2*AD^2= AB^2 + AC^2 - (1/2)*BC^2

2*BE^2= AB^2 + BC^2 - (1/2)*AC^2

Adding the three we get:

2*(CF^2 + AD^2 + BE^2) = AC^2+ BC^2 - (1/2)*AB^2 + AB^2+ AC^2 - (1/2)*BC^2 + AB^2+ BC^2 - (1/2)*AC^2

adding and subtracting the common terms

= (3/2) AC^2 + (3/2) BC^2 + (3/2) AB^2

=> 4*(CF^2 + AD^2 + BE^2) = 3*(

Answer 2

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