Question

In figure given, If $LD$ $\perp$ $MN$, then the value of $L{M}^2+D{N}^2-L{N}^2-M{D}^2$ equals

Answer 1

In right-angle triangle $\Delta LDM$,

$L{M}^2=M{D}^2+L{D}^2$ ( Pythagorus Theorem )

$L{M}^2-M{D}^2=L{D}^2$ .................... (1)

In right-angle triangle $\Delta LDN$ ,

$L{N}^2=L{D}^2+D{N}^2$

$L{N}^2-D{N}^2=L{D}^2$ ..................... (2)

Subtract (2) from (1) ,

$L{M}^2-M{D}^2-(L{N}^2-D{N}^2)=0$

$L{M}^2-M{D}^2-L{N}^2+D{N}^2=0$