Geometric Construction Problem: Why are the segments congruent? The angle $\angle{EDF}$ is constructed from the given angle $\angle{BAC}$ by drawing a set of dotted arcs. It can be determined from the construction that:

$$\overline{AB} \cong \overline{DE}$$
$$\overline{AC} \cong \overline{DF}$$
$$\overline{BC} \cong \overline{EF}$$
Why are the line segments above congruent?
I can't figure out why the three segments are congruent. My initial answer was triangles $BAC$ and $EDF$ are congruent by SSS. But My answer is wrong.
 A: Because you constructed them to be equal.
....
Okay, the assumption is that you have the vertex $A$ and the angle but you do not have the points $B,C$ or $D,E,F$ or the second angle.
Now you are give that task of constructing a second angle that is congruent to the first.  You do the following:
1) You pick a point anywhere.  You call that point $D$.  That will be the vertex of the congruent angle.
2) You make a "base ray" for you angle.  You take a straightedge and draw a line through $D$ in any direction.
3) You go back to your first angle and go to its "base ray" and pick any point on it.  Call that point $C$.
4) You take a compass (you know we talk about compasses and straightedges and compass and straightedge constructions but do we still use and teach them-- anyway a compass allows us to replicate any distance and to draw a circle or an arc of that distance)... You take you compass and set its distance replicate the distance of $AC$.
5) Yous swing your compass to draw an arc of radius $AC$ centered at $A$ until it crosses the other ray of the angle.  You note  where the arc crosses the ray and label that point $B$.
We know $AC = AB$ because the compass has been set to the same radius.
6) You take your compass, still set to the distance $AC$, and take it to the line going through $D$.  You center your compass at $D$ and mark off a point on the line.  That point is $AC$ distance away from $D$.   You label that point $F$
So $AC = DF$ because you constructed it that way.
6) You swing your arc widely.  All the points on this arc will be $AC = AB = DF$ away from $D$.  Somewhere, but you don't know where, on this arc will be one point that will make an angle congruent to $\angle BAC$.
7) Unset your compass.  Put the base of your compass on point $C$ set the distance on you compass to match the distance from $C$ to $B$.  That is the distance $BC$.
8) Place the base of your compass at $F$.  Mark of a point on the arc you made in step 6.  Label that point $E$.
$DE=AB$ because $E$ is one the arc you made in step 6.
$EF=BC$ because your compass was set to $BC$ when you marked off point $E$.
9) Use your straight edge to construct the ray ${AB}^\rightarrow$.
You are done.  
$AB= DE$,$AC=DF$ and $BC= EF$ so $\triangle ABC = \triangle DEF$ and $\angle BAC \cong \angle EDF$.
