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

There is a square $$ABCD$$, line $$EF$$ and point $$G$$ on a plain. 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.

Here is my construction.

# Complexity: 42

• Create a point $$H$$ on line $$AC$$.
• Create 2x2 grid: ACBDI BHCDJ BCDHK AJBCL AKCDM ABHLN ADHMO CDINP BCIOQ
• Create 2x4 grid: ADNQR BCNOS BCOPT ADPQU NPRSV NPTUW
• Create line $$Id\bot EF$$: IOEFX ABVXY CDWXZ EFYZa ABIab AIObc ADNcd
• Create rectangle $$IhGl$$: GNIOe GPIOf NfPeg GgIOh GOINi GQINj OjQik GkINl
• Create $$Gp\bot EF$$: IGhlm GkIdn IOmno EFGop

• 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.
• That's a lot of points. It isn't easy to see what you did, but if there is such a construction not using compasses I'd think it was shorter. Jun 23 at 10:33