Question

5 marks question........

Answer 1

$\huge\textbf{Answer}$


AD is perpendicular to BC. So, /_ADC and /_ADB = 90°

In ∆ ADC

AC² = AD² + CD²

[By Pythagoras Theorem]

AD² = AC² - CD² ____(1)

In ∆ ABD

AB² = AD² + BD²

AD² = AB² - BD² _____(2)

From (1) and (2)

AC² - CD² = AB² - BD²

AC² + BD² = AB² + CD²

AB² + CD² = AC² + BD²

Hence proved.

Answer 2

$\textbf{Answer is in Attachment !}$