Question
In the given figure, if $AD \bot BC$, prove that $A B^2+C D^2=B D^2+A C^2$.
✓
Answer
In right angled $\triangle BDA$,
By pythagoras theorem
$AB^2 = AD^2 + BD^2 ...(i)$
And in right angled $\triangle CDA$ ,
By pythagoras theorem
$AC^2 = CD^2 + AD^2 ...(ii)$
On subtracting Eq$(ii)$ from Eq$(i)$ , we get
$A B^2-A C^2=\left[A D^2+B D^2\right]-\left[C D^2+A D^2\right] $
$A B^2-A C^2=A D^2+B D^2-C D^2-A D^2 $
$A B^2-A C^2=B D^2-C D^2$
$\therefore AB^2+ CD^2= BD^2+ AC^2$
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