Correct option is (B) \(2AD^2 + 2BD^2\)
\(\because\) D is mid-point of BC.
\(\therefore\) BD = CD
Construction :-
Draw \(AE\bot BC\)
In right \(\triangle AED,\)
\(AD^2=AE^2+DE^2\) _______________(1)
In right \(\triangle AEB,\)
\(AB^2=BE^2+AE^2\) _______________(2)
In right \(\triangle AEC,\)
\(AC^2=AE^2+CE^2\) _______________(3)
Now, \(AB^2+AC^2\) \(=(BE^2+AE^2)+(AE^2+CE^2)\) (From (2) & (3))
\(=2AE^2+BE^2+CE^2\)
\(=2(AD^2-DE)^2+BE^2+CE^2\) (From (1))
\(=2AD^2+BE^2+CE^2-2DE^2\)
\(=2AD^2+(BD-DE)^2+(CD+DE)^2-2DE^2\) \((\because BE=BD-DE\;\&\;CE=CD+DE)\)
\(=2AD^2+BD^2+DE^2-2BD.DE\) \(+CD^2+DE^2+2CD.DE-2DE^2\)
\(=2AD^2+BD^2-2BD.DE+BD^2+2BD.DE\) \((\because CD=BD)\)
\(=2AD^2+2BD^2\)
\(=2(AD^2+BD^2)\)
Hence, \(AB^2+AC^2=2AD^2+2BD^2\)