Question
ABCD is a rectangle. Points M and N are on BD such that $\text{AM}\perp\text{BD}$ and $\text{CN}\perp\text{BD}.$ Prove that $BM^2 + BN^2= DM^2+ DN^2$.
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Answer
Given: A rectangle ABCD where $\text{AM}\perp\text{BD}$ and $\text{CN}\perp\text{BD}.$
To prove: $BM^2 + BN^2 = DM^2 + DN^2$
Proof:
Apply Pythagoras Theorem in $\triangle AMB$ and $\triangle CND$,
$A B^2=A M^2+M B^2$
$C D^2=C N^2+N D^2$
Since $A B=C D, A M^2+M B^2=C N^2+N D^2$
$\Rightarrow AM^2-CN^2=ND^2-MB^2 \ldots \text { (i) }$
Again apply Pythagoras Theorem in $\triangle AMD$ and $\triangle CNB$,
$A D^2=A M^2+M D^2$
$C B^2=C N^2+N B^2$
Since $A D=B C, A M^2+M D^2=C N^2+N B^2$
$\Rightarrow AM^2-CN^2=NB^2-MD^2 \ldots$
Equating (i) and (ii),
$N D^2-M B^2=N B^2-M D^2$
l.e., $BM ^2+ BN ^2= DM ^2+ DN ^2$
This proves the given relation.
To prove: $BM^2 + BN^2 = DM^2 + DN^2$
Proof:
Apply Pythagoras Theorem in $\triangle AMB$ and $\triangle CND$,
$A B^2=A M^2+M B^2$
$C D^2=C N^2+N D^2$
Since $A B=C D, A M^2+M B^2=C N^2+N D^2$
$\Rightarrow AM^2-CN^2=ND^2-MB^2 \ldots \text { (i) }$
Again apply Pythagoras Theorem in $\triangle AMD$ and $\triangle CNB$,
$A D^2=A M^2+M D^2$
$C B^2=C N^2+N B^2$
Since $A D=B C, A M^2+M D^2=C N^2+N B^2$
$\Rightarrow AM^2-CN^2=NB^2-MD^2 \ldots$
Equating (i) and (ii),
$N D^2-M B^2=N B^2-M D^2$
l.e., $BM ^2+ BN ^2= DM ^2+ DN ^2$
This proves the given relation.
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