\magnification=\magstep1
\leftline{Comments on Question III A}
\bigskip
\noindent
A.  This is a clever idea, but there is a {\it logical problem}
with this approach.  There is nothing in this proof
that shows that {\it 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 {\it 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 {\it every parallelogram}.

\bigskip
\noindent
There are two ways to fix this:
\bigskip
\item{1)}  Show that every parallelogram comes from 
some rectangle under this construction using bisection
of a pair of parallel sides.  
\item{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.  
\bigskip
\noindent
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.  
\vfill\eject
\leftline{Comments on Question III B, C, and D}
\bigskip
\noindent
B.  Your approach to question B is more seriously flawed because
with this construction you are {\it always constructing
isosceles triangles}.  The idea of the problem was to show
the area formula for {\it an arbitrary triangle}. So you 
need to do something different here.  Here is a better plan:
\bigskip
\item{1)}  You want to start 
from a {\it given, arbitrary triangle} rather than starting 
from the rectangle.
\item{2)}  From the triangle construct either a rectangle
with the twice the area, or a parallelogram with twice
the area.  
\item{3)}  Then apply the given fact for rectangles, or
the result from part A as appropriate.  
\bigskip
\noindent
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 {\it given}.  
\bigskip
\noindent
D.  The intersection points of the inscribed circle and the sides
of a triangle {\it 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 {\it 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.)
\bye
