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

AD is a median and AE $⊥$ BC

To prove:-

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

Consider $A B^{2} + A C^{2}$

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

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

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

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

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

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

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

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

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