$△ A B C$ is an equilateral triangle such that $A D ⊥ B C,$ then $A D^{2} =$

\frac{3}{2} D C^{2}

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2 D C^{2}

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3 C D^{2}

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4 D C^{2}

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Solution

The correct option is D $3 C D^{2}$
Given, equilateral triangle $A B C$ , $A D ⊥ B C$
Since, it is an equilaateral triangle, the perpendicualr also bisects the side.
Hence, $B D = D C$ or $B D = \frac{1}{2} B C = \frac{1}{2} A B$
Now, In $△ A B D$ , we have
$A B^{2} = A D^{2} + B D^{2}$
$\Rightarrow A B^{2} - B D^{2} = A D^{2}$
$\Rightarrow B C^{2} - B D^{2} = A D^{2}$
$\Rightarrow (2 C D)^{2} - C D^{2} = A D^{2}$
$\Rightarrow 3 C D^{2} = A D^{2}$

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