In Traingle ABC, if AD is the median, then show that AB^2 + AC^2 = 2(AD^2+BD^2)

Draw AE perpendicular to BC.

Since angle AED = 90⁰,

Therefore, in triangle ADE

 angle ADE < 90⁰ and angle ADB > 90⁰

Thus, triangle ABD is obtuse angled triangle and triangle ADC is acute angled triangle.

ΔABD is obtuse angled at D and AE is perpendicular to BD produced,

Therefore,

AB² = AD² + BD² + 2 BD x DE ……………(i)

Triangle ACD is acute angled at D and AE is perpendicular to BD produced,

Therefore,

AC² = AD² + DC² - 2 DC x DE

 AC² = AD² + BD² - 2 BD x DE  ………(ii)        (since CD = BD)

Adding (i) and (ii)

AB² + AC² = 2AD² + 2BD²

AB² + AC² = 2(AD² + BD²)