Question

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

• A. $AB\times BC=AD\times DC$
• B. ${AD}^2$ + ${CD}^2$ = ${BD}^2$ + ${AB}^2$
• C. ${AB}^2$ + ${CD}^2$ = ${BC}^2$ + ${AD}^2$
• D. ${AB}^2+{CD}^2={BC}^2\times {AD}^2$

Answer 1

The correct option is C

${AB}^2$ + ${CD}^2$ = ${BC}^2$ + ${AD}^2$

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

${AB}^2={AD}^2+{BD}^2...\left(1\right)$

${BC}^2={CD}^2+{BD}^2...\left(2\right)$

Subracting (2) from (1), we get,

${AB}^2={AD}^2+{BD}^2\underset{–}{{BC}^2={CD}^2+{BD}^2(-)(-)(-)}{AB}^2-{BC}^2={AD}^2-{CD}^2$

Rearranging, we get,
$\Rightarrow {AB}^2+{CD}^2={AD}^2+{BC}^2$