The point $P\left( x,y \right)$ is equidistant from the points $Q\left( c+d,d-c \right)$ and $R\left( c-d,c+d \right)$ then

Point $P(x,y)$ is the equidistant from $Q(c+d,d-x)$ and $R(c-d,c+d)$

$\therefore PQ=PR$

${ PQ }^{ 2 }={ PR }^{ 2 }$

${ \left[ x-\left( c+d \right) \right] }^{ 2 }+{ \left[ y-\left( d-c \right) \right] }^{ 2 }={ \left[ x-\left( c-d \right) \right] }^{ 2 }+{ \left[ y-\left( c+d \right) \right] }^{ 2 }$

${ x }^{ 2 }+{ \left( c+d \right) }^{ 2 }-2.x\left( c+d \right) +{ y }^{ 2 }+{ \left( d-c \right) }^{ 2 }-2.y(d-c)={ x }^{ 2 }+{ \left( c-d \right) }^{ 2 }-2.x.\left( c-d \right) +{ y }^{ 2 }+{ \left( c+d \right) }^{ 2 }+2.y(c+d)$

Or, $x(c+d)+y(d-c)=x(c-d)+y(c+d)$

Or, $(c+d)(x-y)=(c-d)(x+y)$

Or, $cx-cy+dc-dy=cx+cy-dx-dy$

Or, $2dx=2cy$

Or, $dx=cy$