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\centerline{MONT 108N -- Mathematics Through Time}
\centerline{Selected Solutions -- Problem Set 4}
\centerline{November 14, 2010}
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\noindent
II. C)  Method 1:  Let $A_1A_2 \cdots A_n$ be a convex
polygon with $n$ sides and $n$ vertices, where the vertices
$A_i$ are numbered clockwise around the perimeter, and the
edges are $A_1A_2$, $A_2A_3, \ldots, A_{n-1}A_n$ and $A_nA_1$.
Subdivide the polygon into triangles by introducing new
straight lines $A_1A_3, A_1A_4, \ldots, A_1A_{n-1}$ using
Postulate 1 for each line.  There are
$n - 3$ such lines and the original polygon is subdivided
into $n - 2$ triangles (since there is one triangle on
each side of each of the new lines).  We claim that the
sum of the angles in the $n - 2$ triangles equals the
sum of the interior angles in the original polygon.
The angles at $A_2$ and $A_n$ in the first and
last triangles equal the corresponding angles in the original
polygon.  The angle at $A_3$ in the polygon is
$\angle A_2 A_3 A_4$, which equals the sum
$$\angle A_2 A_3 A_1 + \angle A_1 A_3 A_4.$$
Similarly, for each $i$, $i = 3, \ldots, n - 2$,
$$\angle A_{i-1} A_i A_{i+1} = \angle A_{i-1} A_i A_1 + \angle A_1 A_i A_{i+1}$$
which is the sum of angles in the triangles.  Finally, the angle
at $A_1$ in the polygon is
$$\angle A_n A_1 A_2 = \angle A_n A_1 A_{n-1} + \angle A_{n-1}A_1A_{n-2} + \cdots + \angle A_3A_1A_2.$$
Hence the sum of the angles in the $n - 2$ triangles equals the sum of the
angles in the original polygon.  By part B, this means the sum of the angles
in the polygon is $(n-2)$ straight angles.
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\noindent
Method 2 (proof by induction):  The base case for the induction is the case
of a triangle.  By part B, the sum of the angles is one straight angle, which
is $n - 2 = 3 - 2$ straight angles for a figure with $n = 3$ sides.
So the formula works in that case.  Now
suppose the result has been proved for all convex polygons with $k$ sides
and consider a polygon $A_1 A_2 \cdots A_{k+1}$ with $k+1$ sides.  We
number the vertices in the same fashion as in the other proof.  If we
connect $A_1$ and $A_k$ with a straight line (Postulate 1), then the $(k+1)$-sided
polygon is subdivided into a triangle $\triangle A_1 A_k A_{k+1}$ and
a $k$-sided polygon.  By induction, the sum of the angles in the triangle
and the $k$-sided polygon is $1 + (k-2) = (k+1) - 2$ straight angles.
And as in the first proof, the angles at $A_1$ and $A_k$ in the two smaller
polygons sum to the corresponding angle in the larger polygon.  Therefore
we have shown that the sum of the angles in any convex $n$-sided polygon
is $n-2$ straight angles.
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\noindent
III.  A)  Let $ABCD$ be a parallelogram.  Using Postulate 2 extend the side $AB$ to
a straight line so that applying Proposition 12 to drop perpendiculars from $C$ and $D$,
we obtain points $F$ and $G$ on the straight line containing $AB$.
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\put(40,60){\line(0,-1){40}}
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\put(40,23){\line(1,0){3}}
\put(43,23){\line(0,-1){3}}
\put(100,23){\line(1,0){3}}
\put(103,23){\line(0,-1){3}}
\put(15,5){$A$}
\put(35,5){$F$}
\put(75,5){$B$}
\put(95,5){$G$}
\put(35,65){$C$}
\put(95,65){$D$}
\put(115,5){$E$}
\end{picture}
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Proposition 34 implies that in the parallelogram $ABCD$,
$AC = BD$.  By the definition of a parallelogram, $AC$
is parallel to $BD$.  Then Proposition 29 implies that
$\angle CAF = \angle DBG$.  We also have
$\angle CFA = \angle DGB$ since both are right
angles (Postulate 4).  Hence, by Proposition 26 (the AAS
congruence criterion), $\triangle ACF$ and $\triangle BDG$
are congruent, and hence have the same area.  Moreover,
since $AF = BG$ in these congruent triangles, the
Common Notions imply that $CD = FG$.
Now consider the quadrilateral $FGCD$.  This is
a rectangle since $CF$ and $DG$ are parallel (Proposition 28),
and all the interior angles are right angles (Propsition 29).
Moreover $ABCD$ and $FGCD$ have the same area since
the parallelogram is $ACF + CFBD$ and the rectangle
is $CFBD + BDG$ (Common Notion 2).  Therefore the
area of the parallelogram is equal to $(FG)(CF) = (AB)(CF)$
which is what we wanted to show.
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\noindent
B)  Given a triangle $\triangle ABC$.
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\put(40,60){\line(1,0){80}}
\put(80,20){\line(1,2){20}}
\put(15,5){$A$}
\put(75,5){$B$}
\put(35,65){$C$}
\put(95,65){$D$}
\put(115,65){$E$}
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Using Proposition 31 (and Postulate 2), construct straight line $CE$ parallel to $AB$.  Then use
Proposition 3 to mark off $CD$ equal to $AB$ along line $CE$.  Then
join $BD$ using Postulate 1.
Since $CD = AB$ along parallel lines, Proposition 33 implies that
$DB$ = $CA$ and they are parallel too.  Thus $ABCD$ is a parallelogram
by definition.  Now since $CB$ is a straight line falling on parallel
lines, $\angle ABC = \angle DCB$.  Since $AB = CD$ and the side $BC$
is shared, this shows $\triangle ABC$ and  $\triangle BCD$ are congruent
(Proposition 4).  Therefore, the area of $\triangle ABC$ is half
the area of the parallelogram $ABCD$, which is $AB$ times the
altitude (the perpendicular distance from $C$ to $AB$ as
in the previous part).
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\noindent
C.  In the figure for part B, suppose the angle at A is actually
a right angle.  Then we are in the special case in which the altitude equals the side
$AC$.  Then part B implies the area is $\frac{1}{2}(AB)(AC)$.
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