In $△$ ABC, AD is a median. Prove that ${A B}^{2} + {A C}^{2} = 2 ({A D}^{2} + {D C}^{2}) .$

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Solution

In right triangles, $A E B$ and $A E C,$ using Pythagoras theorem

$A B^{2} + A C^{2} = B E^{2} + A E^{2} + E C^{2} + A E^{2} = 2 A E^{2} + (B D - E D)^{2} + (E D + D C)^{2}$

$= 2 A E^{2} + 2 E D^{2} + B D^{2} + D C^{2}$

$B D = D C$

$A B^{2} + A C^{2} = 2 A E^{2} + 2 E D^{2} + 2 B D^{2}$ [since $B D = D C$ ]

$= 2 (A E^{2} + E D^{2} + B D^{2})$

$= 2 (A D^{2} + B D^{2})$

[Using Pythagoras theorem in $A E D$ ]

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