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

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Solution

$\Rightarrow$ $A D$ is a median of $△ A B C .$

$\Rightarrow$ Draw $A E ⊥ B C$ .
In right angled $△ A E B,$

$\Rightarrow$ $(A B)^{2} = (A E)^{2} + (B E)^{2}$ [ By Pythagoeas theorem ] --- ( 1 )
In right angled $△ A C E,$

$\Rightarrow$ $(A C)^{2} = (A E)^{2} + (E C)^{2}$ [ By Pythagoeas theorem ] ---- ( 2 )
Adding ( 1 ) and ( 2 ),

$\Rightarrow$ $(A B)^{2} + (A C)^{2} = (A E)^{2} + (B E)^{2} + (A E)^{2} + (E C)^{2}$

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

$\Rightarrow$ $(A B)^{2} + (A C)^{2} = 2 (A E)^{2} + (B D)^{2} - 2 B D . E D + (E D)^{2} + (E D)^{2} + 2 E D . D C + (D C)^{2}$
$\Rightarrow$ $(A B)^{2} + (A C)^{2} = 2 (A E)^{2} + 2 (E D)^{2} + (B D)^{2} + (D C)^{2}$ [ Since, $B D = D C$ ]

$\Rightarrow$ $(A B)^{2} + (A C)^{2} = 2 (A E)^{2} + 2 (E D)^{2} + 2 (B D)^{2}$ [ Since, $B D = D C$ ]

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

$\Rightarrow$ $(A B)^{2} + (A C)^{2} = 2 [(A D)^{2} + (B D)^{2}]$

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