Proposition 11
To draw a straight line at right angles to a given straight line from a given point on it.
[Desmos graphs are located here]
We will start with a line segment AB, and let C be the given point on it.
We will draw a line from point C at right angles from the segment AB.
We begin by plotting a point D on AC.
Let E be the point on CB such that CE is equal to CD. (Proposition 3 states that we can cut off a line segment that is equal to CD.)
On DE we can construct the equilateral triangle FDE (Proposition 1 shows the construction that shows an equilateral triangle can be built on any line segment), and join points F and C as shown.
We will now show that FC forms right angles with AB at point C by looking at triangles DCF and ECF.
- DC is equal to EC by the way we chose point E.
- CF is common to both triangles.
- DF is equal to FE because they are both sides of the equilateral triangle DFE.
Therefore by Proposition 8 (if two triangles have three equal sides then the angles between two equal sides are equal), the angles DCF and ECF are equal to each other.
One of Euclid’s definitions will help us here: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the angles is right.
The two adjacent angles DCF and ECF are formed when a straight line (CF) is set up on a straight line (AB) and are equal to each other, so each of the angles is right.
This proposition will be used in the next proposition, a construction of drawing a line through a point that is perpendicular to a given straight line.
Questions, comments, or feedback? Please leave a comment below – George
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