If in $△ A B C, A D$ is median and $A E ⊥ B C$ , then prove that $A B^{2} + A C^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}$ .

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Solution

To prove $A B^{2} + A C^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}$

Conditions $A E ⊥ B C$ is drawn

Proof: In $△ A B E$
$A B^{2} = A E^{2} + B E^{2}$ by
pythagoras thorem
(or) $A B^{2} = A D^{2} - D E^{2} + (B D - D E)^{2}$
$= A D^{2} - D E^{2} + B D^{2} + D E^{2} - 2 B D . D E$

$∴ A B^{2} = A D^{2} + B D^{2} - 2 B D \times D E . . . . (1)$
In $△ A E C, A C^{2} = A E^{2} + E C^{2}$
$= (A D^{2} - E D^{2}) + (E D + D C)^{2}$

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

Adding equation $(1)$ and $(2)$ we get
$A B + A C^{2} = 2 A D^{2} + 2 B D^{2} (∵ B D = D C)$
$= 2 A D^{2} + 2 {(\frac{1}{2} B C)}^{2}$

$= 2 A D^{2} + \frac{1}{2} B C^{2}$ as $B D = 1 / 2 B C$

Hence proved.

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Δ A B C

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{A B}^{2} + {A C}^{2} = 2 {A D}^{2} + \frac{1}{2} {B C}^{2}

A D

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△

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

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