In a acute triangle ABC, AD …
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Gaurav Seth 6 years, 10 months ago
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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