Question

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Answer 1

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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{ADC}=\angle{ADB}∠ADC=∠ADB , 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)

\because∵ 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{ADB}\neq\angle{ADC}∠ADB≠∠ADC 
Let 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].