Question
In $\triangle A B C$, $A D$ is perpendicular to $B C$. Prove that: $A B^2+C D^2=A C^2+B D^2$
✓
Answer
Since $\triangle ABD$ and $ACD$ are right triangle right$-$angled at $D,$
$AB^2 = AD^2 + BD^2 ....(i)$
$AC^2= AD^2 + CD^2 ....(ii)$
Subtracting $(ii)$ and $(i),$
we get
$AB^2 - AC^2 = BD^2- CD^2$
$\Rightarrow AB^2 + CD^2 = AC^2 + BD^2.$
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