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Rectangles and Defect Recall that, given a triangle )ABC, the angle sum for the triangle is the sum of the interior angles of )ABC : . Similarly, for a convex quadrilateral the angle sum is the sum of the interior angles of the quadrilateral: Definition: Given a triangle )ABC, the defect difference between 180 and the angle sum of )ABC: is the . For a convex quadrilateral the defect is . Note: The definition of angle measurement (as always non-negative) and the Saccheri-Legendre Theorem combine to insure that for any triangle )ABC, and . Lemma: ~ If )ABC is a triangle and E is a point such that B-E-C, then . A very similar proof can be used to establish: Lemma: If is a convex quadrilateral, then . ~ Exercise. We abbreviate these facts by saying that defect is additive or talking about the additivity of defect. Definition: A rectangle is a convex quadrilateral each of whose angles is a right angle. Note that by definition, the defect of a rectangle is 0. Finally, one lemma before we state and prove the main theorem: Lemma: If is any triangle, then at least two of the interior angles are acute. If the acute angles have vertices A and B, then the foot F of the perpendicular from C to is such that A*F*B. ~ The first statement follows immediately from Saccheri-Legendre. For the second statement, note that F cannot be equal to A or B, since that would make the angles at A or B right angles, contrary to the assumption. Now either F*A*B, F*B*A, or A*F*B. If F*A*B, then pCAB is an exterior angle to and so must be larger in measure than pCFA, which is a right angle. This contradicts our assumption that has acute angles at A. A similar proof shows that A*B*F leads to a contradiction. Thus we must have A*F*B. Big Theorem: The following are equivalent: 1. 2. 3. 4. 5. 6. There is a triangle with defect 0 There is a right triangle with defect 0 There is a rectangle There are arbitrarily large rectangles The defect of every right triangle is 0 The defect of every triangle is 0 Outline of proof: 1 ! 2): Let *()ABC) = 0, and WLOG let be the longest side. By the lemma, there is a point D with A*D*B, such that D is the foot of the perpendicular to from D. Then )ADC and )BDC are right triangles. If *()ABC) = *()ADC) + *()BDC) = 0, then both *()ADC) and *()BDC) must equal 0. 2 ! 3): Let )ABC be right with the right angle at B, and let *()ABC)=0. Then :(pA) + :(pC )= 90. Construct . Then :(pDAC) = 90 - :(pBAC) = :(pBCA), so by SAS, . Then CPCF gives that GABCD is a rectangle. 3 ! 4): We show that we can make a rectangle with dimensions larger than any given N > 0, essentially by pasting together copies of a given rectangle. Start with a given rectangle GABCD. Create points E and F with A*B* E, D*C*F, AB = BE, and DC = CF. By SAS, so :p1 = :p2 and AC = EC by CPCF. We can check that E is interior to pBCF, so that :p1 + :p3 = 90 = :p2 + :p4 implies :p3 = :p4. We can then use SAS to get . Now *()ABC) + *()ACD) = *(GABCD) = 0, so )ABC has defect 0, and :pBAC + :p1 = 90. Because :pBAC + :pCAD = 90 as well, :pCAD = :p1. Similarly, we can show :pBAC = :p3. Thus by ASA, . Now since , CPCF gives us that :pCFE = 90. In an analogous way we can show that pBEF has measure 90, and so GADFE is a rectangle with twice the length of the original. Having shown we can construct such a rectangle, we can apply the process repeatedly on the length and width, and since for any N > 0 there is an M with we can repeat the process to obtain an arbitrarily large rectangle 4 ! 5): Given a right triangle )ABC with right angle at C, create a rectangle whose length and width are both larger than the legs of )ABC. Then, find points on two adjacent sides of the rectangle such that their distances from the common vertex equal the length of the legs. Essentially, we are copying the right triangle into the corner of a larger rectangle. To simplify notation, we will keep the vertex lables A, B, and C for the triangle as shown below. Now *(GCDEF) = 0 since it is a rectangle, and by additivity of defect we must have and . Since and . Finally, since , we must have last!) 5 ! 6): , and (at . Given a triangle )ABC, drop a perpendicular to the longest side (WLOG ), so the foot of the perpendicular is D with A*D*B. This creates two right triangles )DAC and )DBC . By additivity of defect, the defect of )ABC must be 0. 6 ! 1): Trivial. Note that since we know that the EPP is equivalent to every triangle having angle sum 180 (or, alternately, every triangle having defect 0), this theorem adds five more statements to our list of equivalents to the EPP, the most imporant one being: Corollary: The Euclidean Parallel Postulate is equivalent to the existence of a rectangle. Thus, if the Euclidean parallel postulate is not true there can be no rectangles. We use this fact to prove: The All-Or-Nothing Theorem (AKA the Ado Annie Theorem): If there is one line l0 and one exterior point P0 with more than one line through P0 parallel to l0 , then it is true of every line l and exterior point P. ~ Since there is one line l0 and one exterior point P0 with more than one line through P0 parallel to l0 , the EPP fails and we know there cannot be any rectangles. Now, let l be a line and P a point not on the line. In the usual way, construct the usual parallel line: Drop a perpendicular from P to a point Q on line l. Construct a line m perpendicular to through point P. By the Saccheri-Legendre Theorem, m must be parallel to l. Now, through a point R on l, construct a line t perpendicular to l and drop a perpendicular from P to t, with the foot being point S. Now consider GPQRS. The angles at Q, R, and S are all right. Since no rectangle can exist, the angle at P cannot be right, so that But neither can intersect l or it would violate Saccheri-Legendre (or you can apply the Alternate Interior Angles Theorem in Absolute Geometry with t being the transversal). Thus there are two lines parallel to l through P. . The upshot of all this is that the decision on parallelism is a choice between two worlds – Euclidean and Hyperbolic. You have to jump all the way into one or the other. Note that all we have done so far in this set of notes has been done in Absolute Geometry. We haven’t taken a stand yet, but have just explored the consequences of taking such a stand.