Question

in a triangle ABC if AD is a median then show that AB^2+AC^2=2(AD^2+BD^2)

Answer 1

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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.!