Prove every angle has a bisector.

Question

Prove every angle has a bisector.

I have successfully constructed a bisector and justified by construction. Now I need to put it in proof form.

However, I technically do not know midpoints and isosceles triangles yet to prove it. So how would I do this proof without using midpoints and isosceles triangles?

Answer 1

Let $\;O\;$ be the intersection point of two rays forming angle $\;\theta\;$ . Put a compass on this point and draw a circle, which intersect the rays in points $\;A,B\;$ .

Now put compass on point $\;A\;$ and draw an arc between both rays , and without changing the compass do the same when the compass is pinned on point $\;B\;$ and intersecting the first arc , say at point $\;\;$ . These two arcs don't have to be the same measure as in the very first circle on $\;O\;$, but it must be the same opening for the compass on both points $\;A,B\;$!).

Now join with a ruler points $\;O,C\;$ : that's the angle's bisector. Why?

The quadrilateral $\;OACB\;$ is a kite and $\;OC\;$ is the kite's main diagonal which intersects the corresponding angles.

Answer 2

If the measure of the angle can be divided by two, then there is a bisector. Since any number can be divided by two, then any angle has a bisector.

Edit: I found another question similar to this one that you asked, but it was about midpoints and line segments. You could just apply the same thing written here.