Constructing perpendicular with caliper and straightedge

Question

You can only measure length of some segment with caliper and then trace it along already drawn line.

How to construct a perpendicular from a given point to a given line only with caliper straightedge?

My work.

I already know how to build a line, parallel to given one, through a given point, and how to find a midpoint of a given segment, so you can use this constructs in your proofs.

Parallel line

$l$ - given line and $B$ - given point.Let's choose random points $A$ and $M$, then mark point $D$, so $AM=DM$. Construct random point $E$ on the ray $AB$. Construct $EM$ and $BD$ with intersection at point $O$. Construct $AO$ and $ED$ with intersection at point $C$.

Let's suppose that point $C'\in ED$, so $BC' \parallel AD$ and then prove that $C = C'$. In trapezoid $ABC'D$ point of intersection of diagonals, point of intersection of lateral sides, and midpoints of bases lay on one line - $EM$, so $OM$ is point of intersection of diagonals, $AC=AC'$, $C=C'$ and $BC \parallel AD$.

Midpoint

$AB$ is given segment. Construct random $AC$ and mark point $D$, so $AC=CD$. Build $CM \parallel BD$. $M$ is the midpoint of $AB$.

Answer 1

First case: if point $P$ lies outside line $AB$, here's how to construct $H$ on $AB$ such that $PH\perp AB$.

We can suppose WLOG that $PA>PB$ (if $PA=PB$ then $H$ is the midpoint of $AB$). Construct $C$ on ray $AP$ such that $AC=AP+BP$. Construct $D$ on ray $AB$ such that $AD=AP-BP$. Construct $E$ on ray $AB$ such that $AE=2AB$. Join $CE$ and construct the line through $D$ parallel to $CE$, intersecting line $AP$ at $F$.

It is easy to show that $$ AF={AP^2-BP^2\over2AB}. $$ But, on the other hand: $$ AH={AB\over2}+{AP^2-BP^2\over2AB}. $$ Hence, if $M$ is the midpoint of $AB$, we have $MH=AF$ and point $H$ can thus be constructed.

Second case: if point $P$ lies on line $AB$, then take any point $Q$ outside $AB$ and construct $H$ on $AB$, as shown above, such that $QH\perp AB$. Construct then the midpoint $M$ of $PQ$ and point $H'$, symmetric of $H$ with respect to $M$. Line $PH'$ is then the required perpendicular, because $QHPH'$ is a parallelogram.