Math 3181 Name: Dr. Franz Rothe February 25, 2014 All3181\3181_spr14h3.tex Homework has to be turned in this handout. The homework can be done in groups up to three due March 11/12 3 10 10 10 10 10 10 10 10 10 90 Homework 10 Problem 3.1. Give a detailed definition what is meant by congruence of two generic triangles. Moreover, explain with exact notation what the congruence 4ABC ∼ = 0 0 0 4A B C means,—illustrated with a colored figure. 1 Figure 1: How to simply get ASA congruence Proposition 1 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles with a pair of congruent sides, and pairwise congruent adjacent angles are congruent. 10 Problem 3.2. State the ASA congruence, with the notation from the figure above. Prove the theorem. 2 Proposition 2 (SAS Congruence). [Theorem 12 in Hilbert] Given are two triangles. We assume that two sides and the angle between these sides are congruent to the corresponding pieces of the second triangle. Then the two triangles are congruent. 10 Problem 3.3. Complete the following short proof of the SAS-congruence Proposition 2. Short proof of SAS-congruence Proposition 2. Given are two triangles ? . We assume that the angles at A and A0 as well as two pairs of adjacent sides are matched: AB ∼ = A0 B 0 , AC ∼ = A0 C 0 , ∠BAC ∼ = ∠B 0 A0 C 0 By axiom ? we conclude ∠ABC ∼ = ∠A0 B 0 C 0 Now the ? -congruence stated in ? triangle congruence 4ABC ∼ = 4A0 B 0 C 0 to be shown. above yields the Question. We see from Proposition 2 that Hilbert’s SAS-axiom (III.5) implies the full SAS congruence theorem. Why is Hilbert’s Axiom (III.5) weaker than the SAS congruence theorem? 3 Figure 2: Transfer of a triangle. Proposition 3 (Transfer of a triangle). Any given triangle 4ABC can be transferred into any given half-plane H0 of any given ray r, such that one obtains a congruent −−→ triangle 4A0 B 0 C 0 lying in the prescribed half-plane, and the given ray is r = A0 B 0 , emanates from vertex A0 and lies on the side A0 B 0 . 10 Problem 3.4. Use the notation from the figure above. Prove the Proposition 3, starting from the axioms and ASA-congruence. State clearly which axioms you use, and where one has to use the ASA-congruence. 4 Proposition 4 (Extended ASA-Congruence). Given is a triangle, a segment congruent to one of its sides, and a half-plane H0 bounded by this segment . The two angles at the vertices of this side are transferred to the endpoints of the segment, and reproduced in the same half plane. Then the newly constructed rays intersect, and one gets a second congruent triangle. Figure 3: Extended ASA congruence 10 Problem 3.5. Complete the following short proof of the extended ASA-congruence Proposition 4. Short proof of the extended ASA-congruence Proposition 4. Given are the triangle ? −−→ and the segment A0 B 0 ∼ = AB. We define the ray r = A0 B 0 . By Proposition 3 about ? , we transfer the triangle 4ABC into the given half-plane 0 H . We obtain a congruent triangle 4A0 B 0 C 0 with vertex C 0 lying in the prescribed half-plane. Question. In which point does the extended ASA congruence theorem extend the usual ASA congruence theorem? 5 Proposition 5 (The diagonals of the rhombus bisect each other perpendicularly). Given are four different points A, B, C, D lying in one plane such that the segments AB ∼ = BC ∼ = CD ∼ = DA are congruent. Then the segments AC and BD bisect each other perpendicularly at their common midpoint. Figure 4: The diagonals of the rhombus bisect each other. Question. Why do points A and C lie on different sides of line BD, points B and D lie on different sides of the other diagonal AC. Question. Give a reason why the left- and right triangles 4ABD ∼ = 4CBD are congruent (see third figure). 6 Furthermore, the latter two triangles are both isosceles. Hence 4ABD ∼ = 4CDB ∼ holds, too, and ∠ABD = ∠CDB (see first figure in the second row). Similarly, we prove ∠CAB ∼ = ∠ACD. The two segments AC and BD intersect. Let point M be their intersection point. Question. How does one get the congruence 4M AB ∼ = 4M CD (see second figure in the second row). Why do the diagonals AC and BD bisect each other (see last figure in the second row). Question. Justify the triangle congruence 4M AB ∼ = 4M CB. Question. Why are the diagonals AC and BD are perpendicular to each other. 7 10 Problem 3.6. Give purely geometric definitions for supplementary angles and for vertical angles. (Do not use any measurements!) 10 Problem 3.7. Give purely geometric definitions for right angle, acute angle and for obtuse angle. (Use only comparison of angles, not any measurements!) 8 Figure 5: Four right triangles yield a kite. When is it even a rhombus? Proposition 6 (The Hypothenuse Leg Theorem). Two right triangles for which the two hypothenuse, and one pair of legs are congruent, are congruent. Question. Of which one of SAS, SSA, SAA, ASA, SSS congruence contains the hypothenuseleg theorem as a special case? Is there a corresponding unrestricted congruence theorem? 10 Problem 3.8. Complete the following proof of the hypothenuse leg theorem. First proof of the hypothenuse-leg theorem 6 . Given are two right triangles 4ABC and 4A0 B 0 C 0 . As usual, we put the right angles at vertices C and C 0 . We assume congruence of the hypothenuses AB ∼ = A0 B 0 and of one pair of legs AC ∼ = A0 C 0 . We build a kite out of two copies of triangle 4ABC and two copies of triangle 4A0 B 0 C 0 . To this end, one ? into the opposite half plane and gets by transfer of triangle (see Proposition 3) the congruence 4C 0 A0 B 0 ∼ = 4CAB 00 . Especially (3.1) B0C 0 ∼ = B 00 C 9 Since all right angles are congruent (see Proclus’s Proposition), the two supplementary right angles at vertex C imply that points ? lie on a line. −→ Next we transfer segment CA onto the ray opposite to CA and get point D such that askCA ∼ = CD. We produce the two triangles to the right, as shown in the second row of the figure on page 9. Using the right angle, ? imply ∼ 4ABC = 4DBC and ? . Since AB ∼ = A0 B 0 ∼ = AB 00 and DB ∼ = AB ∼ = AB 00 ∼ = DB 00 , one has constructed a symmetric kite with the four points X := A, Y := D, Z1 := B, Z2 := B 00 We can apply Hilbert’s kite-theorem 7 and conclude ∠DAB = ∠XY Z1 ∼ = ∠XY Z2 = ∠DAB 00 The congruence of the same angles is ∠CAB = ∠CAB 00 . Hence SAS congruence implies now ? and hence (3.2) BC ∼ = B 00 C Together, the formulas (3.1) and (3.2) imply B 0 C 0 ∼ = BC. Since all right angles are congruent, a final SAS congruence using the right angles implies congruence 4ABC ∼ = 4A0 B 0 C 0 of the originally given triangles, as to be shown. 10 Proposition 7 (The Kite-Theorem). [Theorem 17 in Hilbert] Let Z1 and Z2 be two points on different sides of line XY , and assume that XZ1 ∼ = XZ2 and Y Z1 ∼ = Y Z2 . ∼ Then the two triangles 4XY Z1 = 4XY Z2 are congruent. 10 Problem 3.9. Complete the proof of the kite-theorem and provide drawings for all three cases. Proof. The congruence of the ? angles of the ? triangle ∼ ∼ 4XZ1 Z2 yields ∠XZ1 Z2 = ∠XZ2 Z1 . Similarly, one gets ∠Y Z1 Z2 = ∠Y Z2 Z1 . Now ? (or substraction) implies (*) ∠XZ1 Y ∼ = ∠XZ2 Y −−−→ Angle addition is needed in case ray Z1 Z2 lies inside the angle ∠XZ1 Y , angle ? −−−→ in case ray Z1 Z2 lies ? angle ∠XZ1 Y . In the special case that either point X or Y lies on the line Z1 Z2 , one gets the same conclusion even easier. One now applies SAS congruence to 4XZ1 Y and ? . Indeed the angles at Z1 and Z2 and the adjacent sides are pairwise congruent. Hence the assertion 4XY Z1 ∼ = 4XY Z2 follows. 11

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