MONT 108N – Mathematics Through Time Selected Solutions – Problem Set 4 November 14, 2010

II. C) Method 1: Let A_{1}A_{2} ⋯ 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_{1}A_{2}, A_{2}A_{3}, …, A_{n−1}A_{n} and A_{n}A_{1}. Subdivide the polygon into triangles by introducing new straight lines A_{1}A_{3}, A_{1}A_{4}, …, A_{1}A_{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 ∠A_{2}A_{3}A_{4}, which equals the sum

∠A_{2}A_{3}A_{1} + ∠A_{1}A_{3}A_{4}.

Similarly, for each i, i = 3, …, n − 2,

∠A_{i−1}A_{i}A_{i+1} = ∠A_{i−1}A_{i}A_{1} + ∠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

∠A_{n}A_{1}A_{2} = ∠A_{n}A_{1}A_{n−1} + ∠A_{n−1}A_{1}A_{n−2} + ⋯ + ∠A_{3}A_{1}A_{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.

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} ⋯ 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 △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.

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.

[figure: A; F; B; G; C; D; E]

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 ∠CAF = ∠DBG. We also have ∠CF A = ∠DGB since both are right angles (Postulate 4). Hence, by Proposition 26 (the AAS congruence criterion), △ACF and △BDG are congruent, and hence have the same area. Moreover, since AF = BG in these congruent triangles, the Common Notions imply that CD = F G. Now consider the quadrilateral F GCD. 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 F GCD have the same area since the parallelogram is ACF + CF BD and the rectangle is CF BD + BDG (Common Notion 2). Therefore the area of the parallelogram is equal to (F G)(CF) = (AB)(CF) which is what we wanted to show.

B) Given a triangle △ABC.

[figure: A; B; C; D; E]

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, ∠ABC = ∠DCB. Since AB = CD and the side BC is shared, this shows △ABC and △BCD are congruent (Proposition 4). Therefore, the area of △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).

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).