Questions: The steps to construct a triangle with side lengths equal to AB using a compass and a straightedge are shown below. Prove that the construction results in an equilateral triangle. Use the drop-down menus to complete the proof. Step 1 Step 2 Step 3 Set the compass to the length of AB and draw a circle centered at point A. Without changing the compass opening, draw a circle centered at point B. Label one of the points of intersection of the two circles C and use a straightedge to draw AC and BC. AC congruent to AB because all radii of circle A are congruent. AC congruent to BC sides are congruent.

The steps to construct a triangle with side lengths equal to AB using a compass and a straightedge are shown below.

Prove that the construction results in an equilateral triangle.
Use the drop-down menus to complete the proof.
Step 1 Step 2 Step 3
Set the compass to the length of AB and draw a circle centered at point A. Without changing the compass opening, draw a circle centered at point B. Label one of the points of intersection of the two circles C and use a straightedge to draw AC and BC.

AC congruent to AB because all radii of circle A are congruent.

AC congruent to BC sides are congruent.

Transcript text: The steps to construct a triangle with side lengths equal to $A B$ using a compass and a straightedge are shown below. Prove that the construction results in an equilateral triangle. Use the drop-down menus to complete the proof. \begin{tabular}{|c|c|c|} \hline Step 1 & Step 2 & Step 3 \\ \hline \begin{tabular}{l} Set the compass to the length of \\ $A B$ and draw a circle centered \\ at point $A$. \end{tabular} & \begin{tabular}{l} Without changing the compass \\ opening, draw a circle centered \\ at point $B$. \end{tabular} & \begin{tabular}{l} Label one of the points of \\ intersection of the two circles \\ $C$ and use a straightedge to \\ draw $A C$ and $B C$. \end{tabular} \\ \hline \end{tabular} \[ \begin{array}{ll} \overline{A C} \cong \overline{A B} \text { because all } & \text { radii of circle } A \end{array} \quad \text { are congruent. } \] \[ \overline{A C} \cong \overline{B C} \] sides are congruent.

Solution

Solution Steps

To prove that the construction results in an equilateral triangle, we need to show that all three sides of the triangle are congruent. The steps involve using the properties of circles and the congruence of radii.

Draw a circle centered at point $ A $ with radius $ AB $.
Draw another circle centered at point $ B $ with the same radius $ AB $.
The intersection of these two circles will give us point $ C $.
By construction, $ AC $ and $ BC $ are radii of the circles centered at $ A $ and $ B $ respectively, and hence are congruent to $ AB $.

Thus, $ AC = AB $ and $ BC = AB $, making $ \triangle ABC $ an equilateral triangle.

Step 1: Draw Circle Centered at $ A $ with Radius $ AB $

Set the compass to the length of $ AB $ and draw a circle centered at point $ A $.

Step 2: Draw Circle Centered at $ B $ with Radius $ AB $

Without changing the compass opening, draw a circle centered at point $ B $.

Step 3: Identify Intersection Point $ C $

Label one of the points of intersection of the two circles as $ C $ and use a straightedge to draw $ AC $ and $ BC $.

Step 4: Prove $ \overline{AC} \cong \overline{AB} $

Since $ AC $ is a radius of the circle centered at $ A $, and all radii of a circle are congruent, we have: \[ \overline{AC} \cong \overline{AB} \]

Step 5: Prove $ \overline{BC} \cong \overline{AB} $

Similarly, since $ BC $ is a radius of the circle centered at $ B $, and all radii of a circle are congruent, we have: \[ \overline{BC} \cong \overline{AB} \]

Step 6: Conclude $ \triangle ABC $ is Equilateral

By the transitive property of congruence, if $ \overline{AC} \cong \overline{AB} $ and $ \overline{BC} \cong \overline{AB} $, then: \[ \overline{AC} \cong \overline{BC} \] Thus, all three sides of $ \triangle ABC $ are congruent, making it an equilateral triangle.

Final Answer

$\boxed{\triangle ABC \text{ is equilateral}}$

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