I was given steps to construct a figure:
1.) Construct a horizontal ray AB and a segment AC at an angle to the ray. Locate point D anywhere on ray AB and construct the segment CD.
2.) Construct the angular bisectors of angle ADC and ACD.
3.) Construct the intersection of E of the two bisectors.
4.) Construct the perpendicular of ray AB to point E.
5.) Construct the intersection of F of the perpendicular and AB. Construct the segment EF and then erase the two angular bisectors.
6.) Construct a circle with center E passing through F (this circle should appear tangent to both rays and CD)
The point E is moving along a certain entity, which in fact is the bisector of angle BAC. Prove that this is indeed so, i.e. that the point E lies on the bisector of angle BAC.
I constructed the figure, I want to make sure it is correct.
Also, I'm not entirely sure how to prove that E lies on the bisector of angle BAC. Any ideas would be appreciated!
You are essentially asking for a proof that the three angle bisectors of a triangle are concurrent. The proof is easy.
The key idea is that a point lies on the bisector of $\angle D$ iff it is equidistant from the lines $AD,CD$. So by construction the point $E$ is equidistant from the lines $AD,CD$ and also equidistant from the lines $AC,CD$.
Hence it is also equidistant from the lines $AC,AD$. Hence it lies on the bisector of $\angle CAD$, otherwise known as $\angle BAC$.