Question

ABCD is a rectangle. Point M or N are on BD such that $AM\perp BD$ and $CN\perp BD.$ Prove that ${BM}^2+{BN}^2={DM}^2+{DN}^2.$

Answer 1

In $△ABM$ and $△CDN$

$\angle AMB=\angle CND$ [ Each ${90}^o$ ]

$AB=CD$ [ Opposite sides of rectangle are equal ]

$\angle ABM=\angle CDN$ [ BD is diagonal, so each ${45}^o$ ]

Thus, $△ABM\cong △CDN$ [ By ASA criteria ]

So, $BM=DN$ [ Corresponding sides of congruent triangle ]

$\Rightarrow$ $B{M}^2=D{N}^2$ ---- ( 1 )

In $△ADM$ and $△BCN$

$\angle AMD=\angle CNB$ [ Each ${90}^o$ ]

$AD=BC$ [ Opposite sides of rectangle are equal ]

$\angle ADM=\angle CBN$ [BD is diagonal, so each ${45}^o$ ]

Thus, $△ADM\cong △CBN$ [ By ASA criteria ]

So, $BN=DM$ [ Corresponding sides of congruent triangle ]

$\Rightarrow$ $B{N}^2=D{M}^2$ ------ ( 2 )

Adding equation ( 1 ) and ( 2 ), we have,

$B{M}^2+B{N}^2=D{M}^2+D{N}^2$

Hence proved