Question

In ΔABC shown below, ∠BAC is congruent to ∠BCA:

Answer 1

Refer to the image attached.

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

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

Construction: Construct a perpendicular bisector from point B to Line segment AC.

Consider triangle BAD and BCD,

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

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

(By the definition of a perpendicular bisector)

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

Therefore, $\Delta ABD \cong \Delta BDC$ by Angle Side Angle(ASA) Postulate.

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

Answer 2

Answer:

Step-by-step explanation:

Consider the triangle ABC, 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.

then, from ΔBAD and ΔBCD, we have

∠BDA=∠BDC(each90)

AD=DA (Property of perpendicular bisector)

∠BAC=∠ACB(given)

Thus, by ASA rule,

ΔBAD ≅ ΔBCD.

Therefore, by CPCTC, AB=BC.

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 of a perpendicular bisector.

ΔBAD is congruent to ΔBCD by ASA rule. Line segment AB is congruent to Line segment BC because corresponding parts of congruent triangles are congruent.(CPCTC)

Consequently, ΔABC is isosceles by definition of an isosceles triangle.