Question

In right angled triangle BAC, ∠BAC = 90°, segments AD, BE and CF are mediuns.
Prove that 2 ( AD² + BE² + CF²) = 3BC².

Answer 1

Answer:

Draw right angle triangle ABC,

draw medians.  AD, BE, CF.  Now Join DF.

Since D and F are midpoints of sides AB and BC,  DF will be parallel to AC and is equal to 1/2 AC.

ADF, ABE, AFC are all  right angle triangles.

LHS = 2 (AD² +  BE² +  CF² )

       = 2 [ (AF² + DF²) + (AB² + AE²) + (AF² + AC²) ]

    =  2 [ (AB²/4 + AC²/4)  + (AB² + AC²) + (AC²/4 + AB²/4) ]

      = 2 [ BC² /2 +  BC² ]

      = 3 ( BC² ]

Answer 2

Answer:

Draw right angle triangle ABC,

draw medians.  AD, BE, CF.  Now Join DF.

Since D and F are midpoints of sides AB and BC,  DF will be parallel to AC and is equal to 1/2 AC.

ADF, ABE, AFC are all  right angle triangles.

LHS = 2 (AD² +  BE² +  CF² )

        = 2 [ (AF² + DF²) + (AB² + AE²) + (AF² + AC²) ]

     =  2 [ (AB²/4 + AC²/4)  + (AB² + AC²) + (AC²/4 + AB²/4) ]

       = 2 [ BC² /2 +  BC² ]

       = 3 ( BC² ]