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Question
In an acute-angled triangle, express a median in terms of its sides.
Solution
We have,
In ΔABC, AD is a median.
Draw AE ⊥ BC
In ΔAEB, by pythagoras theorem
AB2 = AE2 + BE2
⇒ AB2 = AD2 − DE2 + (BD − DE)2 [By Pythagoras theorem]
⇒ AB2 = AD2 − DE2 + BD2 + DE2 − 2BD × DE
⇒ AB2 = AD2 + BD2 − 2BD × DE
⇒ AB2 = AD2 + `"BC"^2/4` - BC x DE ...(i) [BC = 2BD given]
Again, In ΔAEC, by pythagoras theorem
AC2 = AE2 + EC2
⇒ AC2 = AD2 − DE2 + (DE + CD)2 [By Pythagoras theorem]
⇒AC2 = AD2 + CD2 + 2CD × DE
⇒ AC2 = AD2 + `"BC"^2/4` + BC x DE ...(ii) [BC = 2CD given]
Add equations (i) and (ii)
AB2 + AC2 = 2AD2 + `"BC"^2/2`
⇒ 2AB2 + 2AC2 = 4AD2 + BC2 [Multiply by 2]
⇒4AD2 = 2AB2 + 2AC2 − BC2
⇒ AD2 = `(2"AB"^2+2"AC"^2-"BC"^2)/4`
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