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 pinjaraarifisha

4

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 ]

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Answered by Anonymous

0

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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