Question

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

Answer 1

Refer to the image attached.

Given: $\angle BAC$ and $\angle BCA$ are congruent.

To Prove: $\Delta ABC$ is an isosceles triangle.

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 $\Delta BDA, \Delta BDC$

$\angle BDA = \angle BDC= 90^\circ$

(By the definition of perpendicular bisector)

$AD=DC$ (By the definition of perpendicular bisector)

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

$\angle BAC$ = $\angle BCA$ (given)

So,  $\Delta BDA \cong \Delta BDC$ 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.