Question

in a right angled ∆BAC ,angle BAC =90°,segments AD, BE and CF are the medians . prove that 2(AD square+BE square+CF square )=3BCsquare​

Answer 1

Answer:

First, since there is a right angle at A, the triangle is subtended in a semicircle with diameter BC.  So AD = CD = BD = BC / 2 since these are all radii.

2 ( AD² + BE² + CF² )

= 2 ( ( BC / 2 )² + ( AE² + AB² ) + ( AC² + AF² ) )

= 2 ( BC² / 4 + ( AC / 2 )² + AB² + AC² + ( AB / 2 )² )

= BC² / 2  +  2 ( AB² + AB² / 4 + AC² + AC² / 4 )

= BC² / 2  +  2 ( 5 AB² / 4  + 5 AC² / 4 )

= BC² / 2  +  5 ( AB² + AC² ) /2

= BC² / 2 + 5 BC² / 2

= 6 BC² / 2

= 3 BC²