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

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\xa0it is clearly shown that ABC is a triangle, in which AD is a median on BC.\xa0construction :- draw a line AM perpendicular to BC.\xa0we have to prove : AB² + AC² = 2(AD² + BD²)\xa0proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90°\xa0then, from ∆ADB,\xa0according to Pythagoras theorem,\xa0AB² = AD² + BD² ….. (1)from ∆ADC ,according to Pythagoras theorem,\xa0AC² = AD² + DC² …… (2)\xa0AD is median.\xa0so, BD = DC …….(3)from equations (1) , (2) and (3),\xa0AB² + AC² = AD² + AD² + BD² + BD²\xa0AB² + AC² = 2(AD² + BD²) [hence proved ]case 2 :- when\xa0Let us consider that, ADB is an obtuse angle.\xa0from ∆ABM,\xa0from Pythagoras theorem,\xa0AB² = AM² + BM²\xa0AB² = AM² + (BD + DM)²\xa0AB² = AM² + BD² + DM² + 2BD.DM ……(1)from ∆ACM,\xa0according to Pythagoras theorem,\xa0AC² = AM² + CM²\xa0AC² = AM² + (DC – DM)²\xa0AC² = AM² + DC² + DM² – 2DC.DM ……(2)from equations (1) and (2),\xa0AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM – 2DC.DM\xa0AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM – DC.DM) ………..(3)a/c to question, AD is median on BC.so, BD = DC …..(4)and from ADM,\xa0according to Pythagoras theorem,\xa0AD² = AM² + DM² ……..(5)putting equation (4) and equation (5) in equation (3),\xa0AB² + AC² = 2AD² + 2BD² + 2(BD.DM – BD.DM)AB² + AC² = 2(AD² + BD²) [hence proved].\xa0