Question

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

Given: Base ∠BAC and ∠ACB are congruent.

Prove: ΔABC is an isosceles triangle.

When completed (fill in the blanks), the following paragraph proves that Line segment AB is congruent to Line segment BC making ΔABC 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.
is congruent to Line segment DC by _______1________.
ΔBAD is congruent to ΔBCD by the _______2________.
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.


Angle-Side-Angle (ASA) Postulate
corresponding parts of congruent triangles are congruent (CPCTC)

corresponding parts of congruent triangles are congruent (CPCTC)
Angle-Side-Angle (ASA) Postulate

the definition of a perpendicular bisector
Angle-Side-Angle (ASA) Postulate

corresponding parts of congruent triangles are congruent (CPCTC)
the definition of a perpendicular bisector

Answer 1

Answer: 1. the definition of a perpendicular bisector

2. Angle-Side-Angle (ASA) Postulate

Explanation:

The perpendicular bisector is a line that divides perpendicularly a line segment into two equal parts. And, Each point on the perpendicular bisector is the same distance from each of the endpoints of the original line segment.

That is, if a line BD bisects AC perpendicularly,

Then, AD = DC

Now, According to ASA (Angle-Side-Angle) postulate, if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent to each other.

Here, Given:ABC is triangle such that, Base ∠BAC and ∠ACB are congruent.

Prove: ΔABC is an isosceles triangle.

Here, BD bisects AC perpendicularly at point D.

Now, In Δ BDA and Δ BCA

∠BAD ≅ ∠DCB ( given)

∠BDA ≅ ∠BDC ( Right angles )

AD ≅ DC ( By the definition of perpendicular bisector)

Thus, By ASA postulate of congruence,

Δ BDA ≅ Δ BCA

By CPCTC, AB ≅ BC

⇒ Δ ABC is isosceles by definition of an isosceles triangle.