**Problem IV:** Above a given indefinitely long straight line, to construct a perpendicular.

Construction: Conceive the given straight line as going through points A and B, and a perpendicular is to be constructed above it; make BC equal to AB [along the same line] and from B draw BD, making with AB any angle whatsoever, and locate D on that line so that it equals BA or BC, and draw the line from point D through point C. If in that line CF is made equal to CA, and in the line ABC, CE is made equal to CD, I say joining EF makes it be perpendicular to AB.

This may be van Schooten’s trickiest construction. The key to the proof of correctness is to note that the points A, C and D are all the same distance from B, so they lie on a circle centered at B and with AC as a diameter. This makes angle ADC a right angle. Now you only have to figure out why angle CEF is also a right angle.