Homework 9 Solutions and Test 4 Review

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1 Homework 9 Solutions and Test 4 Review Dr. Holmes May 6, Homework 9 Solutions This is the homework solution set followed by some test review remarks (none of which should be surprising). My proofs have now been added. I confidently expect that I have made some mistakes here: I will check them over again sometime tomorrow, and if you find mistakes (mislabelled angles for example), do not hesitate to me. I have corrected a number of errors (mostly but not all typos) in the original. These are pointed out where they occur. The most serious problem was in theorem (thanks to a student for pointing this out). I thought at first that I had made a fundamental error in strategy, but in fact my concept of the proof was correct, but I mislabelled something. Also I left out a case, which is pointed out. In the uniqueness part of I cannot assume that ray AF intersects BD at F (give the point a different name) and ASA congruence is used, not SAS. In I didnt explain what the point D was, a rather important omission. The rest of the errors are typos: this is not to say that typos are not a serious problem in this kind of text. Thanks to the student who pointed out problems with and and keep the comments coming : Prove existence and uniqueness of angle bisectors using SAS and the Isosceles Triangle Theorem. You should diagram out what I am saying as you study this! Corrected a typo, E is the midpoint of BD not BC. 1

2 Let BAC be an angle. Let D be a point on AC such that AB is congruent to AD (Ruler Postulate). Let E be the midpoint of the segment BD. ABD is isoceles, so ABD is congruent to ADB by the Isosceles Angle Theorem, and ABE is congruent to ADE by SAS so BAE is congruent to DAE. E, being on both BD and DB, is on the same side of line AD as B and on the same side of line AB as D by two applications of the Ray Theorem. So E is in the interior of BAD = BAC. So AE is an angle bisector for BAC. I think this is what the author has in mind but Im not convinced that we aren t really using Betweenness for Rays in the bit where we show that E in in the interior. To show uniqueness, let BAC be given and let D be constructed in the same way as above. Suppose that AF is a bisector for BAC. AF intersects BD at a point G by the Crossbar Theorem (again, I m pretty sure this is what the author has in mind but I don t think it avoids the Betweenness Theorem for Rays). We then have BAG congruent to DAG, by ASA (not SAS), so BG is congruent to DG, so G is the midpoint E of BD, and the bisector AF = AG is the same as the bisector AE there is only one : Prove: If one interior angle of a triangle is right or obtuse then the other two angles are acute. Didn t say what D was in the first version...thought that sentence and didn t write it. Let ABC be a triangle and assume that BAC is right or obtuse (that is, its measure is 90). Choose a point D such that B A D. By the Linear Pair Theorem, the measure of DAC and the measure of BAC add to 180, so by algebra the measure of DAC 90. By the Exterior Angle Theorem, both of the angles ACB and ABC have measure strictly less than that of DAC, so strictly less than 90, so they are acute angles. This is a lovely test question : If ABC is a triangle with ABC congruent to ACB, then AB is congruent to AC. ABC is congruent to ACB by ASA, so AB is congruent ot AC. 2

3 If I asked you this as a test question, I would probably specifically tell you to carefully write down the congruences which support the application of ASA here. Do it when you study this : Prove the AAS congruence condition. The version of this proof posted originally was confused [a typo made it very hard to follow, and I didnt realize that I needed to mention another case]. In all problems, when I ask you to state the theorem I do not mean give an informal English statement. I mean that you should state it in formal geometrical language. If I were to give you this problem, stating it in formal geometric language would probably be part of the instructions. Suppose that ABC is a triangle and DEF is a triangle and BC is congruent to EF and BCA is congruent to EF D and BAC is congruent to EDF. Our goal is to prove ABC = DEF. Choose the point G on F D such that F G is congruent to CA. Then ABC is congruent to GEF (write down the three congruence facts which support this). If D = G (which is what we expect), we have DEF = ABC and we are done. We need to eliminate the other cases F G D and F D G. If F G D, notice that EGF is an external angle for DGE, so should be greater in measure than the remote interior angle EDG = EDF. But these angles are congruent: EDF = BAC by our original assumptions and EGF is congruent to BAC because GEF is congruent to ABC. If F D G the argument goes in the same way with D and G interchanged. In my earlier attempt at writing down this proof I was working from a diagram in my head and got it turned around, an illustration of why one should not do this : To construct a congruent copy of a triangle on a given base. Let ABC be a triangle and let DE be a segment congruent to AB. Our aim is to find a point F such that ABC is congruent to DEF. By the Angle Construction component of the Protractor Postulate, there is a ray AG such that CAB is congruent to GDE. Notice that you can choose G to be in your favorite half plane determined by line 3

4 DE: this is not needed for this proof but is needed for applications of it. Now choose F on DG such that DF is congruent to AC (Ruler Postulate). We now have that ABC is congruent to DEF by SAS : Let A, B, C be three distinct points. Prove that if d(a, B)+d(B, C) = d(a, C) then A, B, C are collinear. There was a typo here but not what I thought: the s are correct, I wrote < when I meant >. Prove the contrapositive. if A, B, C are three noncollinear points then d(a, B)+d(B, C) d(a, C) Suppose that A, B, C are noncollinear. Thus ABC is a triangle. It then follows that d(a, B) + d(b, C) > d(a, C) by the Triangle Inequality, and so d(a, B)+d(B, C) d(a, C). That seems awfully short, but it works : Prove the Corresponding Angles Theorem. I m just making a brief note about this: in the diagram on p. 83. the corresponding angles shown are CBB and C B B. Notice that BB A, congruent to C B B by the Vertical Angles Theorem, is with CBB a pair of alternate interior angles. Then by the Alternate Interior Angles Theorem the two lines are parallel. This always happens: if we have a pair of corresponding angles we can replace one of them with an angle congruent to it by the Vertical Angles Theorem in such a way as to get a pair of alternate interior angles : Converse to Euclid s Fifth Postulate Let L and L be two lines cut by a transversal T. If L and L meet on one side of T, then the measures of the interior angles on that side add to strictly less than 180. Let D be the point of intersection of L and L. Let B, B be the points at which L, L meet T. The interior angles are B BD and BB D. Choose E so that E B D. EBB is exterior to BB D so its measure is strictly greater than the measure of the remote interior angle BB D. Further, the measures of B BD and EBB add to 180 by the Linear Pair Theorem. So µ( B BD)+µ( B BD) < µ( B BE)+µ( B BD) = 180, which shows what we want. 4

5 4.8.1: The defect δ( ABC) is defined as 180 σ( ABC), that is 180 minus the sum of the measures of the interior angles of the triangle. Prove that if ABC is a triangle and E is a point on the interior of BC then δ( ABE) + δ( ACE) = δ( ABC). Proof: δ( ABE) + δ( (ACE)) = (180 (µ( ABE + µ( AEB) + µ( BAE) + (180 (µ( ACE) + µ( AEC) + µ( CAE)) (definition of defect) = 360 (µ( ABC) + µ( AEB) + µ( BAC) + µ( ACB) + µ( AEC) + µ( CAB)) (renamed angles which are in the original triangle) = 360 (σ( ABC) + µ( AEB) + µ( AEC)) (definition of σ) = 360 (σ( ABC) + 180) (Linear Pair theorem: AEB and AEC) are a linear pair.) = 180 σ( ABC) algebra = δ( ABC) definition of δ 2 Test 4 Review Remarks section 3.6: Be able to prove (and use) the Isosceles Triangle Theorem. Understand the definition of congruence and the input and output of the SAS postulate and the ASA theorem. What are the six different congruences implied by ABC = DEF? In any given application of SAS or ASA you should be able to identify the three congruences already known which let you apply the theorem and the three new ones you get. 5

6 section 4.1: The Exterior Angle Theorem and the theorem on existence and uniqueness of perpendiculars both have rather elaborate proofs which I will not ask you to write in full. It is still a good idea to look at the proof of the Exterior Angle Theorem: I might set up part of the proof and ask you to add a step or two. You should be ready to use either of these theorems: know what they say. section 4.2: You should know how to prove ASA, and how to use it. I like theorem (a homework exercise above). I m not likely to ask about AAS. section 4.3: Study the proof of the Scalene Inequality. I give my own proof here: I believe it is similar to theirs. Let ABC be a triangle. Suppose that d(a, B) > d(a, C). Find a point D on AB such that d(a, D) = d(a, C); notice that A D B. ACD is isosceles, so ACD is congruent to ADC. CD is between CA and CB (betweenness for rays) so µ( ACB) = µ( ACD) + µ( DCB) > µ( ACD) = µ( ADC). ADC is an exterior angle for CDB, so µ( ADC) > µ( DBC = ABC) by the Exterior Angle Theorem, so we see that µ( ACB) > µ( ABC). If d(a, C) > d(a, B), the same argument shows that µ( ABC) > µ( ACB). If d(a, C) = d(a, B), the Isosceles Triangle Theorem shows that µ( ACB) = µ( ABC). From this it follows that AB > AC if and only if µ( ACB) > µ( ABC). Here is my proof of the Triangle Inequality. Typo corrected, a fairly obvious minus sign in place of an equals sign. If ABC is a triangle, then AB + BC > [AC. You definitely want to draw the diagram here. Draw it with AC at the base of the triangle. Choose D such that A B D and d(b, D) = d(b, C) (Ruler Postulate stuff). BCD is isosceles, so BDC is congruent to BCD. We 6

7 apply the Scalene Inequality to triangle ACD: opposite AC in this triangle we have angle ADC = BDC = BCD. Opposite side AD we have the angle ACD. CB is between CA and CD because B is between A and D (betweenness theorem for rays). Thus µ( ACD) = µ( ACB) + µ( DCB) (angle addition postulate part of the Protractor Postulate) > µ( DCB) = µ( BDC) = µ( ADC), so by the Scalene Inequality AD > AC. But AD = AB + BC be construction, so we have established AB + BC > AC. section 4.4: Study definition 4.4.1: the definition of an alternate interior angle is rather tricky. You should certainly be able to prove the Alternate Interior Angles Theorem: the proof is quite short. You should be able to prove the existence of parallels. You can look at the construction in the book (which we also did in class); I will insert my own construction using a general angle before Sunday evening. section 4.5: Lemmas 4,5,3 and (and the closely related problem 4.8.1) is fair game. For the homework problem you need to know what the defect of a triangle is, of course. Lemma has a proof worth studying: I m less likely to assign you to write the whole proof than to write down part of it and ask you to fill things in. section 4.7: Here is my proof that the Euclidean Parallel Postulate implies that the sum of the angles of a triangle is 180 here. This is something I may ask you about. I begin with my more general construction of parallel lines. Given a line L and a point P not on line L: Choose a point Q on L and draw line P Q. Choose a point R on L. Let α be the measure of RP Q. By the angle construction postulate component of the Protractor Postulate, there is a unique ray P S with S in the half plane cut by line P Q which does not contain R (S is on the other side of line P Q from R) and such that the measure of SP Q is α. Angles RP Q and SP Q are alternate interior angles in the transversal of L and line P S by P Q, and they are congruent, so lines L and P S are parallel by the Alternate Interior Angle Theorem. 7

8 Now I prove the main result: If the Euclidean Parallel Postulate holds and ABC is a triangle, then σ( ABC) = 180. Choose a point D on the other side of line AB fron C such that the measure of DAB is the same as the measure of ABC (Protractor Postulate: this is the construction I just described, stated more briefly). By the Alternate Interior Angle Theorem, line BC and line AD (typo corrected here from the original) are parallel. Similarly, choose a point E on the other side of AC from B such that EAC is congruent to ACB. Line AE is parallel to line BC by the Alternate Interior Angle Theorem. Now, by the Euclidean Parallel Postulate there is exactly one line parallel to line BC passing through A, so A, D, E are collinear. We claim that D A E. Suppose otherwise: if D and E are on the same side of A, choose a point F such that D A F : F will be on the same side of AB as C (the other side from D) and the same side of AC as B (the other side from E), that is in the interior of BAC. But then AF intersects BC by the Crossbar Theorem, and we know that line AD (which includes AF ) is parallel to line BC, which is a contradiction. I do not believe I noticed that this bit was needed when I did the proof in class... Now µ( DAB) + µ( BAC) = µ( DAC) by the angle addition component of the Protractor Postulate, and µ( DAC) + µ( EAC) = 180 by the Linear Pair Theorem. So 180 = µ( DAB) + µ( BAC) + µ( CAE) = µ( ABC) + µ( CAB) + µ( ACB) = σ( ABC). 8