△ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. W

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$△$ ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. Which of the following is true?

Options:

A .

A B \times B C = A D \times D C

B .

{A D}^{2}

+

{C D}^{2}

=

{B D}^{2}

+

{A B}^{2}

C .

{A B}^{2}

+

{C D}^{2}

=

{B C}^{2}

+

{A D}^{2}

D . None of the above

Answer: Option C
:
C

Since ΔADB and ΔDBC are right triangles, using Pythagoras theorem, we can write:

${A B}^{2} = {A D}^{2} + {B D}^{2} . . . (1)$

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

Subracting (2) from (1), we get,
${A B}^{2} = {A D}^{2} + {B D}^{2} {B C}^{2} = {C D}^{2} + {B D}^{2} (-) (-) (-) - ------------------- - {A B}^{2} - {B C}^{2} = {A D}^{2} - {C D}^{2}$
Rearranging, we get,
$\Rightarrow {A B}^{2} + {C D}^{2} = {A D}^{2} + {B C}^{2}$

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