In $Δ A B C,$ if $A D$ is the median, then show that $A B^{2} + A C^{2} = 2 (A D^{2} + B D^{2})$

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Solution

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

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