Problem 2.1. Complete the following proof of Euclid III.20, referring to the figure on page 1.

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$Math 3181 Dr. Franz Rothe October 30, 2015 All3181\3181_fall15t2.tex 2 Solution of Test Name: Figure 1: Central and circumference angle of a circular arc, both obtained as differences Problem 2.1. Complete the following proof of Euclid III.$

1 Math 3181 Dr. Franz Rothe October 30, 2015 All3181\3181_fall15t2.tex 2 Solution of Test Name: Figure 1: Central and circumference angle of a circular arc, both obtained as differences Problem 2.1. Complete the following proof of Euclid III.20, referring to the figure on page 1. Reason for Euclid III.20. We deal with the central angle ω = AOB and the circumference angle γ = ACB of the arc AB. We use the two isosceles triangles BOC and COA. Each one of them has a pair of congruent base angles, called α and β, respectively. One extends the!! segment CO beyond the center O and let C be any point on this extension. Now we see the exterior angle x = AOC of triangle AOC equals!! 2α. Indeed, we use the congruence of the base angles of the isosceles triangle AOC together with Euclid I.32: The exterior angle of a triangle is the sum of!! the two nonadjacent interior angles. 1

2 Similarly, we see that exterior angle y = BOC of triangle COB is!! 2β. For the case drawn in the figure on page 1, points A and O lie on different sides of line BC. In that case, angle subtraction is used at the vertices C and O. Indeed, the central and circumference angle are both obtained as!! differences. The circumference angle is!! γ = α β and the central angle is!! ω = x y. Together with the equations x = 2α and y = 2β one gets as to be shown. ω = x y = 2α 2β = 2γ Problem 2.2 (Construction with compass and straightedge). Given is a line l and a point not lying on the line. Provide a drawing and explain how the perpendicular is dropped with traditional Euclidean tools, using only two circles. Figure 2: Construction of the perpendicular with just two circles. Answer. We choose any two points A and B on the line l and draw the two circles with these centers through the given point P. The two circles intersect at point P and a second point Q. The line P Q is the perpendicular to the given line l through point P. Remark. The two points may lie on the same or different sides of the perpendicular, the construction works for any choice of the points A B. The two circles may have different radius. In any case, they indeed intersect at two points. 2

3 Problem 2.3 (Construction with with compass and straightedge, in neutral geometry). Give a construction of an equilateral triangle, and an angle of 60 with compass and straightedge in neutral geometry. (a) Do the construction and give a description. (b) What can one say about the angles at the vertices of the triangle, in neutral geometry? Why are extra steps needed for the construction of the 60 angle? (c) At which vertex can you get the angle of 60 even in neutral geometry, nevertheless? (d) Convince yourself once more that all reasoning has been done in neutral geometry. Additional to Hilbert s axioms, which intersection property do you need. Figure 3: The construction of a 60 angle at the center O is possible with straightedge compass in neutral geometry. Answer. (a) Description of the construction. Draw any segment AB. The circles about A through B, and about B through A intersect in two points C and D. We get an equilateral triangle ABC. Now we draw a third circle about C through point A, which passes through point B, too. With the two circles drawn earlier, the third circle has additional intersection points H and F. Finally, we draw the 3

4 segments AF, BH and CD. These are the perpendicular bisectors of the sides of triangle ABC. All three intersect in one point O inside triangle ABC. At vertex O, one gets six congruent angles which add up to 360, hence they are all 60. (b) The angles at the vertices may not be 60. Extra steps are needed, because, in neutral geometry, the angle sum of a triangle may be less than two right angles. All we can conclude about the angles at the vertices A, B, C, is their congruence. This follows because the angles opposite to congruent sides are congruent. Still they may all three be less than 60. (c) The 60 angles appear at the center. At vertex O, one gets six congruent angles which add to 360, hence they are all 60. (d) The circle-circle intersection property is needed. All justifications can be given in neutral geometry. Besides Hilbert s axioms, we only need the circle-circle intersection property. 4

Problem 2.4. Complete the proof of the exterior angle theorem given below. Figure 4: A proof of the exterior angle theorem presupposing bisection of a segment. Proof.

5 Problem 2.4. Complete the proof of the exterior angle theorem given below. Figure 4: A proof of the exterior angle theorem presupposing bisection of a segment. Proof. In the given triangle ABC, we bisect the side AC and let M be the midpoint. 1 We extend the ray BM and construct on the extension a point F such that MB = MF. As one easily checks via SAS congruence, one obtains congruent triangles (2.1) MCB = MAF Indeed, these triangles have congruent!! vertical angles at vertex!! M, and by construction two pairs of congruent sides!! MC = MA and!! MB = MF. From the triangle congruence (2.1) we obtain congruent angles γ := ACB = CAF. We choose any point D on the ray opposite to AB. Since by construction A M C and B M F, all three points C, M and F lie on the!! same side of line AB. Hence the ray AF lies in the!! interior of the angle δ := CAD, which is an exterior angle of triangle ABC. One obtains γ = CAF < δ = CAD. As to be shown, the exterior angle δ is indeed!! greater than the!! nonadjacent interior angle γ. 1 This proof is presupposing a valid construction to bisect a segment. 5

6 Proposition 1 (The Kite-Theorem). [Theorem 17 in Hilbert] Let Z 1 and Z 2 be two points on different sides of line XY, and assume that XZ 1 = XZ2 and Y Z 1 = Y Z2. Then the two triangles XY Z 1 = XY Z2 are congruent. Problem 2.5. Complete the proof of the kite-theorem and provide drawings for all three cases. Proof. The congruence of the!! base angles of the isosceles triangle XZ 1 Z 2 yields XZ 1 Z 2 = XZ2 Z 1. Similarly, one gets Y Z 1 Z 2 = Y Z2 Z 1. By!! angle addition (or substraction, respectively, ) (*) XZ 1 Z 2 = XZ2 Z 1 and Y Z 1 Z 2 = Y Z2 Z 1 imply XZ 1 Y = XZ 2 Y Angle addition is needed in case ray Z 1 Z 2 lies inside the angle XZ 1 Y, angle!! subtraction in case ray Z 1 Z 2 lies!! outside the angle XZ 1 Y. In the special case that either point X or Y lies on the line Z 1 Z 2, one gets the same conclusion even easier. One now applies SAS congruence to XZ 1 Y and!! XZ 2 Y. Indeed the angles at Z 1 and Z 2 and the adjacent sides are pairwise congruent. Hence the assertion XY Z 1 = XY Z 2 follows. Answer. Here are drawings for the three cases. Figure 5: The symmetric kite 6

Figure 6: Where on the boundary can guys A and B meet the quickest? Problem 2.6. Two guys who are equally good runners stand at points A and B on the border of a place with the shape shown in the figure on page 7.

7 Figure 6: Where on the boundary can guys A and B meet the quickest? Problem 2.6. Two guys who are equally good runners stand at points A and B on the border of a place with the shape shown in the figure on page 7. They are allowed to run across the place, but they want to meet at any point on the boundary. Which point do they need to choose to meet as quick as possible. Give a construction and explain your reasoning. 7

The runners pass congruent segments AM and BM until they meet at some point M.

8 Figure 7: They meet at the intersection of the perpendicular bisector of AB with the boundary marked nearest. Answer. The runners pass congruent segments AM and BM until they meet at some point M. Hence they meet where the perpendicular bisector of AB intersects the boundary. There are two such points, of which they choose the nearer one. 8

Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points.

Math 3181 Dr. Franz Rothe Name: All3181\3181_spr13t1.tex 1 Solution of Test I Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points. Definition