Comments on Question III A

A. This is a clever idea, but there is a logical problem with this approach. There is nothing in this proof that shows that every parallelogram can be constructed by the method you are proposing. (That is true, of course, but you have not said or shown why.) So your proof shows in effect that the areas of a particular class of parallelograms (constructed by the method you describe) has the desired area formula. This is not a complete argument since we wanted to establish the formula for every parallelogram.

There are two ways to fix this:

1) Show that every parallelogram comes from some rectangle under this construction using bisection of a pair of parallel sides.

2) “Run your argument backwards.” By this, I mean start from an arbitrary parallelogram and show that if you “cut and paste” a right triangle, you get a rectangle of the same area as the original parallelogram, so the desired formula follows.

If you think about it, these are essentially equivalent. But 2) seems to be a more natural way to think about the proof to me.

Comments on Question III B, C, and D

B. Your approach to question B is more seriously flawed because with this construction you are always constructing isosceles triangles. The idea of the problem was to show the area formula for an arbitrary triangle. So you need to do something different here. Here is a better plan:

1) You want to start from a given, arbitrary triangle rather than starting from the rectangle.

2) From the triangle construct either a rectangle with the twice the area, or a parallelogram with twice the area.

3) Then apply the given fact for rectangles, or the result from part A as appropriate.

C. There is a similar problem here to the problem on B, though it is easier to fix. You want to think of the right triangle as given and either apply part B or else construct a rectangle whose area is twice the area of the given triangle. Don’t bisect a side of an isosceles triangle, because then you need to show that every right triangle comes from that construction. Assume the right triangle is given.

D. The intersection points of the inscribed circle and the sides of a triangle do not necessarily bisect the sides. You should not need to use that to establish the desired area formula either. (On the other hand, the lines from the center to the vertices of the triangle do necessarily bisect the angles, and that gives a construction of the inscribed circle: Its center is the point where the three angle bisectors meet. For the purposes of this question, though, you can just assume that there is an inscribed circle; you do not need to show how to construct it.)