Let's find the lengths of all sides:
[tex] AB=\sqrt{(4-2)^2+(5-4)^2}=\sqrt{4+1} =\sqrt{5} ,\\ BC=\sqrt{(2-4)^2+(4-3)^2}=\sqrt{4+1} =\sqrt{5},\\ CD=\sqrt{(4-6)^2+(3-4)^2} =\sqrt{4+1} =\sqrt{5},\\ AD=\sqrt{(6-4)^2+(4-5)^2}=\sqrt{4+1} =\sqrt{5} [/tex].
All sides are congruent, then this quadrilateral may occur to be square or rhombus. Check whether sides AB and AD are perpendicular. With this aim you should find vectors [tex] \vec{AB} [/tex] and [tex] \vec{AD} [/tex]:
[tex] \vec{AB}=(2-4,4-5) =(-2,-1)[/tex] ,
[tex] \vec{AD}=(6-4,4-5) =(2,-1)[/tex] .
The dot product: [tex] \vec{AB}\cdot \vec{AD}=-2\cdot 2+(-1)\cdot (-1)=-4+1=-3\neq 0 [/tex]. This means that sides AB and AD are not perpendicular and quarilateral ABCD is not square.
Answer: correct choice is E.