This is a STILL image.

It is NOT a movie.

PS-3

As in the previous two tasks, as point A moves to the right, the angle at A gets smaller so the slope of the bisector of angle A gets closer to the horizontal, ie the line 'dips down'. This might suggest that as A moves to the right, P moves downwards. However, it turns out that this is not the case.

This task is closely related to our second task, where ABCD was a parallelogram. Here we are working with the right-hand portion of the parallelogram, with point P corresponding to point E of the inner shape EFGH. What happens to E when A is dragged along DA produced?

It turns out that E is always midway between the parallel sides AD and BC. In the current task, this means that when A is dragged 'horizontally' along DA produced, P moves horizontally too, and stays the same distance from AD as from BC. One way to prove this is to use the fact that points on the bisector of angle A are equidistant from AD and AB, and that points on the bisector of angle B are equidistant from BA and BC.

Incidentally, the angle made by the bisectors at P is a right angle (as was the case for the angles of EFGH when ABCD was a parallelogram). Again, you might want to ask students to explain why.

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