The common compass and straight-edge construction of a perpendicular to a line $\ell$ from a given point $P$ takes $4$ steps:
$\hspace{10pt}$ $1$. Draw a circle centered at $P$. it will cross $\ell$ at $2$ points.
$\hspace{10pt}$ $2$. From the first intersection point $A$, draw a circle with radius
$\overline{AP}$.
$\hspace{10pt}$ $3$. From the second intersection point $B$, draw a circle with radius
$\overline{AB}$.
$\hspace{10pt}$ $4$. Draw a line from $P$ to the new intersection point between the
previous two circles, it will be the desired perpendicular.
This method can be clearly seen in this diagram:
However, my professor told me there was a way to this in only three steps! I have been trying to figure out for a long time and am beginning to think he is tricking us. I can't find anything online about this supposed $3$ step solution. Does it exist and if so, what is it?
Best Answer
Apparently drawing a circular arc is a step
Pick a point on the line, call it $A.$ Around $A$ draw the circle that passes through $P.$
Pick a second point on the line, call it $B.$ Around $B$ draw the circle that passes through $P.$
THe two circles intersect in $P$ and at a second point $P'$ off the line and the line through $PP'$ is perpendicular to the original line
Here the point $P$ is on the graph paper at $(4,3) $