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C is the incenter of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABC and DBC are congruent?



It is given that C is the incenter of triangle ABD, so segment BC is an altitude of angle ABD.

Angles ABC and DBC are congruent according to the definition of an angle bisector.

Segments AB and DB are congruent by the definition of an isosceles triangle.

Triangles ABC and DBC share side BC, so it is congruent to itself by the reflexive property.

By the SAS postulate, triangles ABC and DBC are congruent.



Triangle ABD with segments BC, DC, and AC drawn from each vertex and meeting at point C inside triangle ABD.

There is an error in line 1; segment BC should be an angle bisector.

The proof is correct.

There is an error in line 3; segments AB and BC are congruent.

There is an error in line 5; the ASA Postulate should be used

Respuesta :

Based on the definition of an angle bisector, there is an erro in line 1, because, BC is an angle bisector.

What is an Angle Bisector?

An angle bisector is a line segment that bisects an angle into congruent parts.

In the proof given, since C is the incenter of the isoscles triangle ABD, therefore, line segment BC bisects ∠ABD.

This implies that BC is an angle bisector of ∠ABD not altitude.

Therefore, based on the definition of an angle bisector, there is an erro in line 1, because, BC is an angle bisector.

Learn more about an angle bisector on:

https://brainly.com/question/24334771

Answer:

the proof is correct

Step-by-step explanation:

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