Proof:
Point D is on a line CD perpendicular to AB, and of course D belongs to AB.
Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).
We end up with 3 equations:
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)
(Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0
Subtitute the first two equations in the third one gives:
(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0
Distributing to solve for t gives:
(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0
which gives:
t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
getting rid of the negative signs:
t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
Once you have t, you can figure out the coordinates for D from the first two equations.
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)