Math, asked by Varun1870, 1 year ago

AD, BE and CF are medians of the triangle ABC, then prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²)

Answers

Answered by apsnaruka1267
9

Answer:


We can prove that if in ABC AD , BE and CF are median and altitude to respective sides , So we get


So,


BD = DC = 12 BC


AE = EC = 12 AC

And

AF = FB = 12 AB


Now we apply Pythagoras theorem in ∆ AFC , As :


AC2 = CF2 + AF2


AC2 = CF2 + ( 12 AB )2 ( As we know AF = 12 AB )


AC2 = CF2 + 14 AB2

Taking LCM we get


4 AC2 = 4CF2 + AB2 ---------------------- ( 1 )


Now we apply Pythagoras theorem in ∆ BCE , As :


BC2 =BE2 + EC2


BC2 = BE2 + ( 12 AC)2 ( As we know EC = 12 AC )


BC2 = BE2 + 14 AC2

Taking LCM we get


4 BC2 = 4BE2 + AC2 ---------------------- ( 2 )

And

Now we apply Pythagoras theorem in ∆ ADB , As :


AB2 = AD2 + BD2


AB2 = AD2 + ( 12 BC )2 ( As we know BD = 12 BC )


AB2 = AD2 + 14 BC2

Taking LCM we get


4 AB2 = 4AD2 + BC2 ---------------------- ( 3 )


Now we add equation 1 , 2 and 3 ,and get


4 AC2 + 4 BC2 + 4 AB2 = 4CF2 + AB2 + 4BE2 + AC2 + 4AD2 + BC2


3 AC2 + 3 BC2 + 3 AB2 = 4CF2 + 4BE2 + 4AD2


3 ( AB2 + BC2 + AC2 ) = 4 ( AD2 + BE2 + CF2 ) ( Hence proved )



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