Question

In a right angled triangle BAC, angle BAC = 90 degree, segments AD, BE and CF are the medians. Prove that 2 (AD square + BE square + CF square) = 3 BC square. ​

Answer 1

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,

=$2( {AD}^{2} + {BE}^{2} + {CF}^{2}$

=$2[( {AF}^{2} + {DF}^{2} ) + ( {AB}^{2} + {AE}^{2} ) + ( {AF}^{2} + {AC}^{2} )]$

=$2(( \frac{ {AB}^{2} }{4} + { \frac{ {AC}^{2} }{4} }) + ( {AB}^{2} + {AC}^{2} ) + ( \frac{ {AC}^{2} }{4} + \frac{ {AB}^{2} }{4} ))$

=$2( \frac{ {bc}^{2}}{2} + {bc}^{2})$

=$3( {bc}^{2} )$

Answer 2

Answer:

Step-by-step explanation:

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,

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