The perpendicular from $A$ on sides $B C$ of triangle intersects $B C$ at $D$ . Such that $D B = 3 C D$ . Prove that $2 A B^{2} = 2 A C^{2} + B C^{2}$ .

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Solution

We have,
$B C = C D + D B$
$\Rightarrow B C = C D + 3 C D (D B = 3 C D)$
$\Rightarrow B C = 4 C D$
$\Rightarrow C D = \frac{B C}{4} - (e q 1)$
$\Rightarrow$ $D B = 3 C D = \frac{3 B C}{4} - e q (2)$

$I n Δ A D C$
$A C^{2} = C D^{2} + A D^{2}$
$\Rightarrow A C^{2} = {(\frac{B C}{4})}^{2} + A D^{2} (f r o m e q 1)$
$\Rightarrow {(A D)}^{2} = A C^{2} - \frac{B C^{2}}{16}$
$\Rightarrow {(A D)}^{2} = \frac{16 A C^{2} - B C^{2}}{16}$

$I n Δ A D B$
$A B^{2} = A D^{2} + D B^{2}$
$\Rightarrow A B^{2} = \frac{16 A C^{2} - B C^{2}}{16} + {(\frac{3 B C}{4})}^{2} (f r o m e q 2)$
$\Rightarrow A B^{2} = \frac{16 A C^{2} - B C^{2} + 9 B C^{2}}{16}$
$\Rightarrow A B^{2} = \frac{16 A C^{2} + 8 B C^{2}}{16}$
$\Rightarrow A B^{2} = \frac{8 (2 A C^{2} + B C^{2})}{16}$
$\Rightarrow 2 A B^{2} = 2 A C^{2} + B C^{2}$
Proved

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