In a triangle, when any two of the altitude, median, perpendicular bisector, and angle bisector from a vertex coincide, the triangle is isosceles

Question

Given a vertex and the corresponding opposite side of a triangle, there are 4 key segments (as I understand it):

(1) altitude (line drawn from vertex perpendicular to opposite side (or an extension of it));

(2) median (line drawn from vertex to midpoint of opposite side);

(3) perpendicular bisector (perpendicular line drawn from midpoint of side---in general does not hit vertex); and

(4) angle bisector (line drawn from vertex to side such that line bisects angle at vertex).

I have seen the following stated:

A) In general, for a given vertex-side pair, each of these lines is different.

B) If any line segment plays at least 2 of the above 4 roles, then the triangle must be isosceles. In other words, whenever the four lines above are not all different, the triangle must be isosceles.

Are (A) and (B) correct? If so, could someone provide a proof of (B)?

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Edit: I see that all 4 roles are played by the same segment if and only if the triangle is isosceles. Rephrasing the question, then, if a segment plays any 2 of the 4 roles, must it play the roles of all 4, i.e. must the triangle be isosceles?

There are six cases:

1. altitude = median

2. altitude = perp. bis.

3. altitude = ang. bis.

4. ang. bis. = perp. bis.

5. median = perp. bis.

6. median = ang. bis.

So is it true that each of the above conditions independently implies that a triangle is isosceles?

3 and 4 is clear to me because of the congruency of the two triangles that results. 1 and 2 and 5 are equivalent; if the altitude also bisects the side, it's the median and the perp. bisector. So, 1 and 6 are not clear to me.