Question

in an acute angled triangle abc ad is the median prove that 4AD^2 + BC^2 =2AB^2+ 2AC^2​

Answer 1

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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 +

BC24 - 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 +

BC24 + BC x DE ...(ii) [BC = 2CD given]

Add equations (i) and (ii)

AB2 + AC2 = 2AD2 +

BC22

⇒ 2AB2 + 2AC2 = 4AD2 + BC2 [Multiply by 2]

⇒4AD2 = 2AB2 + 2AC2 − BC2

Answer 2

ANSWERS:

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