Math, asked by janhavikanire91, 9 months ago

In ∆ABC, angle A=90°, AD,BE,CF are medians prove that 2AB²=2AC²+BC²​

Answers

Answered by radhe00
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 ]

Answered by Anonymous
1

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 ]

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