Question
In figure, if $AD \perp BC,$ then prove that $AB ^2+ CD ^2= BD ^2+ AC ^2$.
✓
Answer
By Pythagoras theorem
$\text { In } \triangle ABD, AB^2=AD^2+BD^2$
$\Rightarrow AD^2=AB^2-BD^2 \ldots \text { (i) }$
$\text { In } \triangle ADC, AC^2=AD^2+CD^2$
$\Rightarrow AD^2=AC^2-CD^2 \ldots \text { (ii) }$
From $(i)$ and $(ii),$
$AB^2-BD^2=AC^2-CD^2$
$\Rightarrow AB^2+CD^2=BD^2+AC^2$
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