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 $\triangle\text{AMB}$ and $\triangle\text{CND},$
$AB^2 = AM^2 + MB^2$
$CD^2 = CN^2 + ND^2$
Since $AB = CD, AM^2 + MB^2 = CN^2 + ND^2$
$\Rightarrow AM^2 - CN^2 = ND^2 − MB^2 … (i)$
Again apply Pythagoras Theorem in $\triangle\text{AMD}$ and $\triangle\text{CNB},$
$AD^2 = AM^2 + MD^2$
$CB^2 = CN^2 + NB^2$
Since $AD = BC, AM^2 + MD^2 = CN^2 + NB^2$
$\Rightarrow AM^2 - CN^2 = NB^2 - MD^2 … (ii)$
Equating (i) and (ii),
$ND^2 - MB^2 = NB^2 - MD^2$
I.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 $\triangle\text{AMB}$ and $\triangle\text{CND},$
$AB^2 = AM^2 + MB^2$
$CD^2 = CN^2 + ND^2$
Since $AB = CD, AM^2 + MB^2 = CN^2 + ND^2$
$\Rightarrow AM^2 - CN^2 = ND^2 − MB^2 … (i)$
Again apply Pythagoras Theorem in $\triangle\text{AMD}$ and $\triangle\text{CNB},$
$AD^2 = AM^2 + MD^2$
$CB^2 = CN^2 + NB^2$
Since $AD = BC, AM^2 + MD^2 = CN^2 + NB^2$
$\Rightarrow AM^2 - CN^2 = NB^2 - MD^2 … (ii)$
Equating (i) and (ii),
$ND^2 - MB^2 = NB^2 - MD^2$
I.e., $BM^2 + BN^2 = DM^2 + DN^2$
This proves the given relation.
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