In ∆ABC shown below, ∡BAC is congruent to ∡BCA. Given: Base ∡BAC and ∡ACB are congruent.
Prove: ∆ABC is an isosceles triangle.

Construct a perpendicular bisector from point B to line segment AC.
Label the point of intersection between this perpendicular bisector and line segment AC as point D.
m∡BDA and m∡BDC is 90° by the definition of a perpendicular bisector.
∡BDA is congruent to ∡BDC by the definition of congruent angles. Line segment AD is congruent to line segment DC by ___1.
∆BAD is congruent to ∆BCD by the _2______. Line segment AB is congruent to line segment BC because corresponding parts of congruent triangles are congruent (CPCTC).
Consequently, ∆ABC is isosceles by definition of an isosceles triangle.

options are:
a)1. Angle-Side-Angle (ASA) Postulate
2. corresponding parts of congruent triangles are congruent (CPCTC)

b) 1. corresponding parts of congruent triangles are congruent (CPCTC)
2. Angle-Side-Angle (ASA) Postulate

c) 1. the definition of a perpendicular bisector
2. Angle-Side-Angle (ASA) Postulate

d) 1. corresponding parts of congruent triangles are congruent (CPCTC)
2. the definition of a perpendicular bisector

Construction: Construct a perpendicular bisector from point B to line segment AC. Label the point of intersection between this perpendicular bisector and line segment AC as point D.

Proof:

Consider [tex]\Delta BDA, \Delta BDC[/tex]

[tex]\angle BDA = \angle BDC= 90^\circ[/tex]

(By the definition of perpendicular bisector)

[tex]AD=DC[/tex] (By the definition of perpendicular bisector)

So, Line segment AD is congruent to DC by the definition of perpendicular bisector.

[tex]\angle BAC[/tex] = [tex]\angle BCA[/tex] (given)

So, [tex]\Delta BDA \cong \Delta BDC[/tex] by ASA congruence postulate.

∆BAD is congruent to ∆BCD by the ASA congruence Postulate.

Line segment AB is congruent to line segment BC because corresponding parts of congruent triangles are congruent (CPCTC).

So, Option C is the correct answer.