Saccheri Quadrilaterals

Transcription

1 Saccheri Quadrilaterals efinition: Let be any line segment, and erect two perpendiculars at the endpoints and. ark off points and on these perpendiculars so that and lie on the same side of the line, and =. Join and. The resulting quadrilateral is a Saccheri Quadrilateral. Side is called the base, and the legs, and side the summit. The angles at and are called the summit angles. Lemma: Saccheri Quadrilateral is convex. ~ y construction, and are on the same side of the line, and by PSP, will be as well. If intersected at a point F, one of the triangles ªF or ªF would have two angles of at least measure 90, a contradiction (one of the linear pair of angles at F must be obtuse or right). Finally, if and met at a point E then ªE would be a triangle with two right angles. Thus and must lie entirely on one side of each other. So, G is convex. F

2 Theorem: The summit angles of a Saccheri Quadrilateral are congruent. ~ The SSS version of using SS to prove the base angles of an isosceles triangle are congruent. G G, by SSS, so p p. orollaries: The diagonals of a Saccheri Quadrilateral are congruent. (Proof: ª ª by SS; PF gives =.) The line joining the midpoints of the base and summit of a quadrilateral is the perpendicular bisector of both the base and summit. (Proof: Let and be the midpoints of summit and base, respectively. Use SSS on G and G. The angles at and are congruent by PF, and form a linear pair, so must have measure 90.) If each of the summit angles of a Saccheri Quadrilateral is a right angle, the quadrilateral is a rectangle, and the summit is congruent to the base. (Proof: onsider diagonal. HL gives ª ª so =.)

3 Lemma: If G has right angles at and, then: < iff μp < μp, = iff μp = μp, and > iff μp > μp. ~ If =, G is Saccheri and we have already shown μp = μp. If <, then let E be a point such that *E* and E =. Then GE is Saccheri and μpe = μpe. E is interior to p, so μp > μpe = μpe. ut μpe > μp by exterior angle inequality, so μp > μp. n exactly analogous proof will establish > implies μp > μp. E We now have one direction of each iff statement. ow suppose μp = μp. If, then < or >. ut we have shown this either gives μp > μp or μp < μp, in contradiction to μp = μp. So μp = μp implies =. Similarly, if μp > μp but Ý, then #, and either μp = μp or μp < μp, again contradicting μp > μp. Similarly, μp < μp must imply <. (ote how we got the converse of each implication since the three implications taken together exhaust all the possibilities.)

4 Theorem: If the summit angles of a Saccheri Quadrilateral are acute, the summit has greater length than the base. ~ Given G, connect the midpoints and of summit and base. onsider G, with right angles at and (we proved this above). y the previous lemma, if p is acute, and therefore less than p in measure, >. Similarly, considering G, if p is acute and therefore less than p in measure, >. ombining these, since ** and **, >. ote: If the summit angles are obtuse, we can just as easily, and in the exact same way, prove that the base is longer than the summit. Since we already know that if the summit angles are right, we have a rectangle, with summit and base of equal length, we can summarize in the following way: If the summit angles of a Saccheri Quadrilateral are: acute, the summit is longer than the base right, the summit is equal to the base and the quadrilateral is a rectangle obtuse, the summit is shorter than the base.

5 In absolute geometry, we cannot establish that the summit angles are in fact right angles. ut we can eliminate the obtuse case, and in the process, get another almost Euclidean result. Theorem: The summit angles of a Saccheri Quadrilateral are either acute or right. ~ egin with Saccheri Quadrilateral G, with right angles at and (so below it is drawn upside down relative to how we have usually drawn it). Extend to a point E with **E such that = E. Let be the midpoint of. onsider ªE and ª. Since E = =, and =, we have ªE ª by SS. This means p pe by PF, and so **E (we have previously proved this). Thus we have a triangle ªE, and pe pe. E ow the sum of the summit angles of G is: μpe + μpe + μp. The sum of the angles of triangle ªE is: μpe + pe + μp. ut these sums are equal, since pe pe. Thus, since the angle sum of the triangle is no more than 180, so is the sum of the summit angles. Since the summit angles are equal in measure, they must be acute. orollary: The length of the summit of a Saccheri Quadrilateral is greater than or equal to the length of the base.

6 This lets us now prove one more interesting thing about Saccheri Quadrilaterals: They are, in fact, parallelograms. efinintion: parallelogram is a convex quadrilateral in which opposite sides are parallel. Theorem: Saccheri Quadrilateral is a parallelogram. Outline of proof: We have already noted that lines and cannot meet (if they met at a point F, then ªF would have two right angles). Suppose and were to meet at a point G. Either **G or G**. WLOG suppose **G. Then since the summit angle at is acute, the angle pg, being supplementary, is obtuse. Similarly, the angle pg is supplementary to the right angle at. Thus ªG has two right or obtuse angles, a contradiction. The case for G** is similar. G

7 Theorem: The line joining the midpoints of two sides of a triangle has length less than or equal to one-half of the third side. (ote: in Euclidean geometry, the inequality is replaced by equality.) Outline of proof: Let ª be given, and locate midpoints and of and, respectively. rop perpendiculars from and to line at and, respectively. Since and lie on opposite sides of from, they lie on the same side of and just as for Saccheri Quadrilaterals we can show G is convex. ' '

8 Our next goal is to show that G is in fact a Saccheri Quadrilateral. To do so, we drop a perpendicular from to at point Q. There are three cases: (1) p and p are both acute; (2) one of p and p is right; and (3) one of p and p is obtuse: In ase 1, **Q**. In ase 2, ==Q, and **. In ase 3, Q*** and **. ' ' Q ase 1 ' ' = = Q ase 2 ' Q ' ase 3

9 We ll do ase I here, and leave the others as exercises. 1 2 ' ' Q ow = and = by construction. Since they are vertical angles, pq p and pq p. y H, ªQ ª and ªQ ª. So Q and is a Saccheri Quadrilateral. ow = Q, and = Q. Since **Q**, = + Q + Q + = 2 Q + 2 Q = 2 (Q + Q) = 2, so = ½(). Since is the base and the summit of Saccheri Quadrilateral G, #. Thus, = ½() # ½().