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Integration Using Substitution

In the follow...

Question

In the following figure, $A D$ is the perpendicular to $B C$ and $D$ divides $B C$ in the ratio $1 : 3.$ Prove that $2 A C^{2} = 2 A B^{2} + B C^{2} .$

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Solution

$B D : D C = 1 : 3 \Rightarrow B D = \frac{1}{4} B C$ and $C D = \frac{3}{4} B C$

$A C^{2} = A D^{2} + C D^{2}$ and $A B^{2} = A D^{2} + B D^{2}$
$\Rightarrow A C^{2} - A B^{2} = C D^{2} - B D^{2}$
$= {(\frac{3}{4} B C)}^{2} - {(\frac{1}{4} B C)}^{2}$

$= \frac{9}{16} B C^{2} - \frac{1}{16} B C^{2}$

$= \frac{8}{16} B C^{2} = \frac{1}{2} B C^{2}$

$∴ 2 A C^{2} - 2 A B^{2} = B C^{2}$

$i . e . 2 A C^{2} = 2 A B^{2} + B C^{2}$

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A D ⊥ B C

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2 A C^{2} = 2 A B^{2} + B C^{2} .

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