Question

in triangle ABC, if AD is the median tgen show that AB^2 +AC^2=2(AD^2+BD^2)

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Answer 1

     

Draw an altitude AM from A to BC at M.

Here, BD = CD.

Assume that AM is at left of AD. See the figure.

So that,

AB² = BM² + AM²

⇒ AB² = (BD - DM)² + AM²

⇒ AB² = BD² - 2•BD•DM + DM² + AM²

⇒ AB² = BD² - 2•BD•DM + AD²   →   (1)

And,

AC² = CM² + AM²

⇒ AC² = (CD + DM)² + AM²

⇒ AC² = CD² + 2•CD•DM + DM² + AM²

⇒ AC² = CD² + 2•CD•DM + AD²  →   (2)

(1) + (2)

AB² + AC² = BD² - 2•BD•DM + AD² +CD² + 2•CD•DM + AD²

⇒ AB² + AC² = 2AD² + CD² + 2DM(CD - BD) + BD²

⇒ AB² + AC² = 2AD² + BD² + 2DM(BD - BD) + BD² [Because BD = CD]

⇒ AB² + AC² = 2AD² + BD² + 0 + BD²

⇒ AB² + AC² = 2AD² + 2BD²

⇒ AB² + AC² = 2(AD² + BD²)

Hence proved!

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Answer 2

Answer:

AB² + AC² = 2(AD² + BD²).

Step-by-step explanation:

(i)

In ΔAED,

⇒ AD² = AE² + DE²

⇒ AE² = AD² - DE²

(ii)

In ΔAEB,

⇒ AB² = AE² + BE²

           = AD² - DE² + BE²

           = AD² - DE² + (BD + DE)²  {BE = BD + DE}

           = AD² - DE² + BD² + DE² + 2BD * DE - DE²

           = AD² + BD² + 2BD * DE

(iii)

In ΔAEC,

⇒ AC² = AE² + EC²

           = AD² - DE² + EC²

           = AD² - DE² + (DC - DE)²

           = AD² - DE² + DC² + DE² - 2DC * DE

           = AD² + BD² - 2BD * DE {DC = BD}

On solving (ii) & (iii), we get

⇒ AB² + AC² = AD² + BD² + 2BD * DE + AD² + BD² - 2BD * DE

                     = AD² + BD² + AD² + BD²

                     = 2(AD² + BD)²

Hence proved.!