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Notes
1) While it is easy to disprove the above proof in Q1-Q9 with reference to the 1st case by noting that the intersection of the angle bisector of ∠C and the perpendicular bisector of AB always falls outside the triangle (on the circumcircle), this is not applicable to the 3rd and 4th cases shown above. In fact the above 'proof' is not valid, since in general for a non-isosceles triangle, one of the points D and F always falls inside the triangle and the other one outside of the triangle. This invalidates the addition of segments used in the 'proof'. (For a more detailed discussion, see Northrop, 1945: 98-102).
'...........However, the proof is completely valid in the 2nd case where the angle bisector of ∠C intersects the opposite side AB at its midpoint E. In this case, we can deduce that ABC is isosceles with CA = CB (in which case the angle bisector of ∠C coincides with the perpendicular bisector of AB).
2) While the main purpose of this activity is to caution students about reasoning with faulty diagrams, a positive spin-off is that it naturally leads to the interesting result that in any non-isosceles triangle the angle bisector of an angle intersects the perpendicular bisector of its opposite side on the circumcircle. A slightly more general formulation of the same result (since it's also valid for an isosceles triangle) is that the angle bisector of a triangle bisects the arc of the circumcircle formed or subtended by its opposide side. This formulation is often useful in other problem solving contexts. See for example, Cyclic Quadrilateral Angle Bisectors Rectangle Result and Cyclic Quadrilateral Midpoints of Arcs Theorem.
3) Many ingenious paradoxes can arise by virtue of construction errors or mistaken assumptions in diagrams. There is often great educational value in using such paradoxes in a mathematics classroom. There are several resources available for teachers to consult and to select from: for example Movshovitz-Hadar & Webb (1998) and an old delighful classic by Northrop (1944).