In a acute triangle ABC, AD is a median. Prove that 4AD^2 + BC^2 = 2AB

1.	In a acute triangle ABC, AD is a median. Prove that 4AD^2 + BC^2 = 2AB^2 + 2AC^2
Answer» \xa0Given:\xa0In ∆ ABC, AD is the medianConstruction:\xa0Draw AE ⊥BC\xa0\xa0Now since AD is the median∴ BD = CD =BC ....... (1)\xa0In ∆ AEDAD2\xa0= AE2\xa0+ DE2 (Pythagoras theorem)⇒ AE2\xa0= AD2\xa0– DE2 ......... (2)\xa0In ∆ AEBAB2\xa0= AE2\xa0+ BE2\xa0= AD2\xa0– DE2\xa0+ BE2\xa0(from (2))= (BD\xa0+ DE)2\xa0+ AD2\xa0– DE2\xa0(∵ BE = BD + DE)= BD2\xa0+ DE2\xa0+ 2BD·DE + AD2\xa0– DE2= BD2\xa0+ AD2\xa0+ 2·BD·DE\xa0In ∆ AEDAC2\xa0= AE2\xa0+ EC2= AD2\xa0– DE2\xa0+ EC2 (from (5))= AD2\xa0– DE2\xa0+ (DC – DE)2= AD2\xa0– DE2\xa0+ DC2\xa0+ DE2\xa0– 2DC·DE= AD2\xa0+ DC2\xa0– 2DC·DE\xa0Adding (3) and (4) we get\xa0AB2\xa0+ AC2\xa0=BC2\xa0+ AD2\xa0+ BC·DE\xa0+ AD2\xa0+BC2\xa0– BC·DE⇒ 2 (AB2\xa0+ AC2) = BC2\xa0+ 4AD2⇒ 4AB2\xa0+ BC2\xa0= 2AB2\xa0+ BC2

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