Question

In the given fig. if AD ⊥ BC. Prove that AB² + CD² = BD² + AC².

Answer 1

$\textbf{Answer:}$

In Fig. ABC is a triangle in which

AD⊥BC

Now we have 2 right-angled triangle.

 ∆ADB and ∆ ADC

In ∆ABD

AD²=AB²-BD² (Pythagoras theorem)

 ∆ACD=AD²-CD²

AD²=AD²

AB²-BD²=AD²-CD²

AB²+CD²=BD²+AD²

Hence proved.


$\textbf{Hope it helps you!}$

Answer 2

By Pythagoras Theorem,



In ∆ ABD,
$\mathbf{AB^{2} = {AD}^{2} + {DB}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$




In ∆ ABC,

$\mathbf{AC^{2}= AD^{2}+CD^{2}} \: \: \: \: \: \: \: \: \: \: \: \: \: ... (ii)$






Now,


$\mathbb{Subtract \: \: ( ii ) \: \: from \: \: \: ( i ), }$


AB² - AC² = AD² + BD² - AD² - CD²


=> AB² - AC² = BD² - CD²


=> AB² + CD² = BD² + AC²








$\textbf{ \: \: Hence, \: Proved}$
.