Question
From given figure, In $\triangle ABC , AD \perp BC$, then prove that $AB ^2+ CD ^2= BD ^2+ AC ^2$ by completing activity.
Activity: From given figure, In $\triangle A C D$, By pythagoras theorem
$ A C^2=A D^2+\square$
$\therefore A D^2=A C^2-C D^2 $
Also, In $\triangle ABD$, by pythagoras theorem,
$ A B^2=\square+B D^2$
$\therefore A D^2=A B^2-B D^2 \quad \ldots . .(I I)$
$\therefore \square-B D^2=A C^2-\square$
$\therefore A B^2+C D^2=A C^2+B D^2 $
Activity: From given figure, In $\triangle A C D$, By pythagoras theorem
$ A C^2=A D^2+\square$
$\therefore A D^2=A C^2-C D^2 $
Also, In $\triangle ABD$, by pythagoras theorem,
$ A B^2=\square+B D^2$
$\therefore A D^2=A B^2-B D^2 \quad \ldots . .(I I)$
$\therefore \square-B D^2=A C^2-\square$
$\therefore A B^2+C D^2=A C^2+B D^2 $
Activity:
