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:

enter image description here

$$\overline{AB} \cong \overline{DE}$$ $$\overline{AC} \cong \overline{DF}$$ $$\overline{BC} \cong \overline{EF}$$

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.

  • 1
    $\begingroup$ I suppose because the radii of the given arcs are the same. $\endgroup$
    – Bernard
    May 4, 2020 at 15:32
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    $\begingroup$ if $AB, AC$ and $DE, DF$ are radii of the congruent circles, they are congruent. Then you can use SAS to determine that the triangles are congruent. $\endgroup$
    – Vasili
    May 4, 2020 at 15:34
  • $\begingroup$ Notice that they used differently dashed lines for arcs $BC$/$EF$ and for the small arcs at $B$/$E$. This is to signify that the first two arcs were drawn with one opening of the compass, and the other two arcs with another (potentially different) opening. $\endgroup$
    – user700480
    May 4, 2020 at 15:38
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    $\begingroup$ The are equal because you constructed them to be equal. $\endgroup$
    – fleablood
    May 4, 2020 at 16:19
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    $\begingroup$ .... however, I have to wonder what the "question" was. Was the op asked why are the equal? Well, there is no reason other than they were constructed to be..... The OP's "answer" that the triangles are congruent by SSS is ... backwards. The triangles are congruent BECAUSE the distances are equal. We can't say the distances are equal because the triangles are congruent... because we don't know the triangles are congruent. $\endgroup$
    – fleablood
    May 4, 2020 at 16:35

1 Answer 1


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 $\angle BAC \cong \angle EDF$.


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