ABCD is a rectangle. Points M and N are on BD such that AM ⊥ BD and CN ⊥ BD.


$A B C D$ is a rectangle. Points $M$ and $N$ are on $B D$ such that $A M \perp B D$ and $C N \perp B D$. Prove that $B M^{2}+B N^{2}=D M^{2}+D N^{2}$.


Given: A rectangle ABCD where AM  BD and CN  BD.

To prove: $\mathrm{BM}^{2}+\mathrm{BN}^{2}=\mathrm{DM}^{2}+\mathrm{DN}^{2}$


Apply Pythagoras Theorem in $\triangle \mathrm{AMB}$ and $\triangle \mathrm{CND}$,



Since $A B=C D, A M^{2}+M B^{2}=C N^{2}+N D^{2}$

$\Rightarrow \mathrm{AM}^{2}-\mathrm{CN}^{2}=\mathrm{ND}^{2}-\mathrm{MB}^{2} \ldots$ (i)

Again apply Pythagoras Theorem in $\triangle A M D$ and $\triangle C N B$,



Since $A D=B C, A M^{2}+M D^{2}=C N^{2}+N B^{2}$

$\Rightarrow \mathrm{AM}^{2}-\mathrm{CN}^{2}=\mathrm{NB}^{2}-\mathrm{MD}^{2} \ldots$ (ii)

Equating (i) and (ii),


I.e., $B M^{2}+B N^{2}=D M^{2}+D N^{2}$

This proves the given relation.


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