Construction of Perpendicular using Rusty Compass
Theorem
Let $AB$ be a line segment.
Using a straightedge and rusty compass, it is possible to construct a straight line at right angles to $AB$ from the endpoint $A$, without extending $AB$ past $A$.
Construction
Mark $AC$ equal to the fixed radius of the rusty compass.
Construct two arcs whose centers are at $A$ and $C$ respectively of radius $AC$.
Let them meet at $D$.
Produce $CD$.
Measure $E$ from $D$ using the rusty compass so that $DE = CD$.
Then $AE$ is the required straight line at right angles to $AB$.
Proof
As $DE = CD = DA$, the points $A$, $C$ and $E$ all lie on a circle of radius $AC$.
$CE$ is a straight line through the centers of circle $ACE$ and so is a diameter of circle $ACE$.
Hence by Thales' Theorem, $\angle CAE$ is a right angle
$\blacksquare$
Historical Note
This construction was discussed by Abu'l-Wafa Al-Buzjani in a work of his from the $10$th century.
Sources
- 1986: J.L. Berggren: Episodes in the Mathematics of Medieval Islam
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Abul Wafa ($\text {940}$ – $\text {998}$): $43$