See figure, here it is clearly shown that ABC is a triangle, in which AD is a median on BC.

construction :- draw a line AM perpendicular to BC.

we have to prove : AB² + AC² = 2(AD² + BD²)

proof :- case 1 :- when angle ADB is equals to angle ADC, it means AD is perpendicular on BC and both angles are right angle e.g., 90°

then, from ∆ADB, according to Pythagoras theorem, AB² = AD² + BD² ..... (1)

from ∆ADC ,according to Pythagoras theorem, AC² = AD² + DC² ...... (2)

AD is median. so, BD = DC .......(3)

from equations (1) , (2) and (3), AB² + AC² = AD² + AD² + BD² + BD² AB² + AC² = 2(AD² + BD²) [hence proved ]

case 2 :- when angle ADC is not equal to angle ADBLet us consider that, ADB is an obtuse angle. from ∆ABM, from Pythagoras theorem, AB² = AM² + BM² AB² = AM² + (BD + DM)² AB² = AM² + BD² + DM² + 2BD.DM ......(1)

from ∆ACM, according to Pythagoras theorem, AC² = AM² + CM² AC² = AM² + (DC - DM)² AC² = AM² + DC² + DM² - 2DC.DM ......(2)

from equations (1) and (2), AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM

AB² + 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, according to Pythagoras theorem, AD² = AM² + DM² ........(5)

putting equation (4) and equation (5) in equation (3),

AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)

AB² + AC² = 2(AD² + BD²) [hence proved].