How do you construct the perpendicular with a square and a straightedge?

Question

There is a square $ABCD$, line $EF$ and point $G$ on a plane. Can you construct the foot of perpendicular to line $EF$ through point $G$, using only a straightedge (in traditional Euclidean constructions)? Is there any construction shorter than mine?

I'd appreciate it if you posted a construction or provided any kinds of reference.

The following construction has complexity 30.

1. Construct 3x3 grid.

   1. Construct the middle point K of line segment AB. You can prove this by Ceva's theorem.
      Choose any point H on line CD. AHBCI ACBHJ ABIJK
      

   2. Simply construct the center of the square and the other middle points of each side.
      ACBDL ACDKM ADBMN CDKLO BCLNP
      

2. Construct the parallel line and the perpendicular line of line EF through point L.

   1. Construct the perpendicular bisector RS of line segment LQ to construct the symmetric line LU of line EF about line RS.
      ACEFQ BQKPR DQNOS EFRST ABLTU
      

   2. Construct the perpendicular line LW of line EF through point L by making it symmetric to LU about LN.
      CUNPV BVCDW ABLWX
      

   3. Construct the line of line EF through point L.
      ACNUY ADKYZ
      

3. Construct the perpendicular line of line EF through G by making c symmetric to G about line LZ.
   GWLZa GXLZb WbXac EFGcd
   

And here is the result. It's really complex.

• The complexity of a construction can be described by the quantity of points.
• XYZUV is short for "creating point $V$ which is the intersection of line $XY$ and $ZU$".
• Notice that points can be in either capital and lowercase letters.