In the figure $A B = B C$ and $A D$ is perpendicular to $C D$ . Prove that:
$A C^{2} = 2 B C . D C$

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Solution

In $△ A D C$ ,
$A C^{2} = A D^{2} + D C^{2}$ ....(1)

In $△ A D B$ ,
$A B^{2} = A D^{2} + D B^{2}$
$∴ A D^{2} = A B^{2} - D B^{2}$ ...(2)

In $△ A D C$ ,
$= (A B^{2} - D B^{2}) + {(D B + B C)}^{2}$ ....from (1) and (2)
$= B C^{2} - D B^{2} + D B^{2} + B C^{2} + 2 D B . B C$ ...(As, $A B = B C$ )
$= 2 B C^{2} + 2 D B . B C$
$= 2 B C (B C + D B)$
$= 2 B C . D C$

Hence proved

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