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?
 1. It is given that C is the incenter of triangle ABD, so segment BC is an altitude of angle ABD.
2. Angles ABC and DBC are congruent according to the definition of an angle bisector.
3. Segments AB and DB are congruent by the definition of an isosceles triangle.
4. Triangles ABC and DBC share side BC, so it is congruent to itself by the reflexive property.
5. By the SAS postulate, triangles ABC and DBC are congruent.

a.) There is an error in line 1; segment BC should be an angle bisector.
b.) The proof is correct.
c.) There is an error in line 3; segments AB and BC are congruent.
d.) There is an error in line 5; the ASA Postulate should be used.

Yes, the proof correctly justifies that triangles ABC and DBC are congruent

Step-by-step explanation:

1) The incenter is given by the intersection of three angle bisectors of the three angles of the triangle

Therefore, given that the line BC connects the vertex B to the incenter, we have that BC is an altitude (gives the height of the angle ∠ABD)

2) The segment BC bisects ∠ABD, being that segment BC is an altitude of the angle and also leads to the incenter of the triangle ABD

Therefore, ∠ABC and ∠DBC are congruent by the definition of bisection of ∠ABD

3) Given that triangle ABD is an isosceles triangle, we have segments AB and DB are congruent

4) Segment BC of triangles ABC and DBC is congruent to itself by reflexive property

5) Therefore, as obtained above, segment AB, ∠ABC, and segment BC of triangle ΔABC are congruent to the corresponding segment BD, ∠DBC, and segment BD of triangle ΔDBC, we have that ΔABC and ΔDBC are congruent by Side-Angle-Side rule of congruency

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

d. vertex

step-by-step explanation: