Question

given: \\( \triangle abc \\) is equiangular
prove: \\( \overline { a b } \cong \overline { b c } \cong \overline { c a } \\)
statements

1. \\( \triangle a b c \\) is equiangular

2 \\( \angle a \cong \angle b \cong \angle c \\)

1. \\( \overline { a b } \cong \overline { c a } \\)
2. \\( \overline { a b } \cong \overline { b c } \\)

5.
reasons

1. given

2.

1. converse to isosceles triangle theorem
2. converse to
3. transitive property of \\( \cong \\)

a. def. equiangular \\( \delta \\)
b. \\( \overline { a b } \cong \overline { b c } \cong \overline { c a } \\)
c. \\( m \angle a + m \angle b + m \angle c = 180 \\)
d. pythagorean theorem
e. \\( m \angle a = m \angle b = m \angle c \\)
f. isosceles triangle theorem

Explanation:

Step1: Definition of equiangular triangle

By the definition of an equiangular triangle, all angles are congruent. So, if \(\triangle ABC\) is equiangular, then \(\angle A\cong\angle B\cong\angle C\). The reason for statement 2 is "Def. equiangular \(\triangle\)" (option a).

Step2: Using the converse of the Isosceles Triangle Theorem

The converse of the Isosceles Triangle Theorem states that if two angles of a triangle are congruent, then the sides opposite those angles are congruent. For \(\overline{AB}\cong\overline{BC}\), we use the same converse of the Isosceles Triangle Theorem (option f).

Step3: Applying the transitive property

The transitive property of \(\cong\) states that if \(a\cong b\) and \(b\cong c\), then \(a\cong c\). Since \(\overline{AB}\cong\overline{CA}\) and \(\overline{AB}\cong\overline{BC}\), by the transitive property, \(\overline{AB}\cong\overline{BC}\cong\overline{CA}\). So, the statement for step 5 is \(\overline{AB}\cong\overline{BC}\cong\overline{CA}\) (option b).

Answer:

1. a. Def. equiangular \(\triangle\)
2. f. Isosceles Triangle Theorem
3. b. \(\overline{AB}\cong\overline{BC}\cong\overline{CA}\)