Proposition 9
To bisect a given rectilinear angle.
[Desmos graphs for this proposition]
This proposition is a construction that shows how to bisect an angle. We begin with the angle BAC.
Choose a point D at random on AB and cut off a segment AE from AC that is equal to AD. (Proposition 3)
Join DE.
Construct the equilateral triangle DEF be constructed using a point F on the opposite side of DE from point A . (Proposition 1)
Let AF be joined.
Now we will show that AF bisects the angle BAC by using the triangles DAF and EAF.
- AD is equal to AE, since AE was cut off AC to equal AD.
- AF is common to both triangles.
- DF is equal to EF, since they are two of the sides of an equilateral triangle.
The conditions for Proposition 8 have been met, so angles DAF and EAF are equal to each other. Therefore the angle BAC has been bisected.
Now that we have learned how to bisect a rectilinear angle, we will learn how to bisect a given straight line (segment) in the next proposition.
As always, questions and feedback is welcome – George