Book 1 – Proposition 10

Proposition 10

To bisect a given straight line.

[Desmos.com graphs for this proposition]

Suppose we start with a line segment AB.

Construct the equilateral triangle ABC on it. (Proposition 1 covers how to construct an equilateral triangle om a line segment.)

Let angle ACB be bisected by the line segment CD, where the point D lies on the line segment AB. (Proposition 9 covers how to bisect a rectilinear angle.)

We will now compare the two triangles ACD and BCD.

AC is equal to BC because they are two sides of the equilateral triangle ABC.
Side CD is common to both triangles.
Angle ACD is equal to angle BCD, since CD bisects the angle ACB.

Since the two triangles have two equal sides, with equal angles between those equal sides, then the two bases AD and BD must be equal. (Proposition 4 states that if two triangles have equal side-angle-side, then the third sides must also be equal.)

Therefore AB has been bisected at D.

If you have any questions or feedback I’d love to hear it. Please leave a comment – George