Question

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.

Answer 1

The incenter C of isosceles triangle ABD suggests that BC is an angle bisector, not an altitude. Given AB=DB and ABC=DBC, by the SAS Postulate, triangles ABC and DBC are congruent.

Step-by-step explanation:

The assertion that C is the incenter of isosceles triangle ABD is crucial to this proof. As the incenter, the point C is indeed the point where the angle bisectors of the angles of triangle ABD intersect. This means that line BC is an angle bisector rather than an altitude. Therefore, line BC bisects the vertex angle ABD into two congruent angles, ABC and DBC.

In an isosceles triangle, such as ABD, we know that two sides are congruent, specifically AB and DB. The SAS postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Here, we have two congruent sides (AB=DB) and a shared side (BC) with the inclusion of congruent angles (ABC=DBC) between those sides. Therefore, by the SAS postulate, triangles ABC and DBC are congruent.