In an acute angled triangle ABC , AD is median in it,prove  

4AD2   +BC2  =2AB2   +2AC2 

 

Given: In ∆ ABC, AD is the median

Construction: Draw AE ⊥BC 

 

Now since AD is the median

∴ BD = CD = BC             ....... (1)

 

In ∆ AED

AD2 = AE2 + DE2                 (Pythagoras theorem)

⇒ AE2 = AD2 – DE2            ......... (2)

 

In ∆ AEB

AB2 = AE2 + BE2 

= AD2 – DE2 + BE2 (from (2))

= (BD + DE)2 + AD2 – DE2 (∵ BE = BD + DE)

= BD2 + DE2 + 2BD·DE  + AD2 – DE2

= BD2 + AD2 + 2·BD·DE

 

In ∆ AED

AC2 = AE2 + EC2

= AD2 – DE2 + EC2                 (from (5))

= AD2 – DE2 + (DC – DE)2

= AD2 – DE2 + DC2 + DE2 – 2DC·DE

= AD2 + DC2 – 2DC·DE

 

Adding (3) and (4) we get

 AB2 + AC2 = BC2 + AD2 + BC·DE + AD2 + BC2 – BC·DE

⇒ 2 (AB2 + AC2) = BC2 + 4AD2

⇒ 4AB2 + BC2 = 2AB2 + BC2

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