In $△ A B C$ , $D$ is mid point of $B C .$ Then,
$A B^{2} + A C^{2} = 2 (. . . . . . . . . . . .)$

C D^{2} + A C^{2}

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A D^{2} + C D^{2}

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A D^{2}

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C D^{2}

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Solution

The correct option is B $A D^{2} + C D^{2}$
$B D = D C$
$A E = E C \Rightarrow A C = \frac{1}{2} E C$ .... $D E | | A B$
$D E = \frac{A B}{2}$
In $△ C E D, D C^{2} = D E^{2} + E C^{2}$
$\Rightarrow 4 D C^{2} = A C^{2} + A B^{2}$
$A B^{2} + A C^{2} = 2 C D^{2} + 2 C D^{2}$
In $△ A D E, A D^{2} = A E^{2} + D E^{2}, A E^{2} + D C^{2} - E C^{2}$
$\Rightarrow A D^{2} = D C^{2}$
$∴ 2 C D^{2} = 2 A D^{2}$
Then, $A B^{2} + A C^{2} = 2 (C D^{2} + A D^{2})$

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Similar questions

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

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

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

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

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If $A D ⊥ B C$ , prove that ${A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}$ . [2 MARKS]

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