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Quadrilaterals Definition: If A, B, C and D are any four points no three of which are collinear. If the points are so situated that no pair of the segments and intersect at an interior point, then the set QABCD = is a quadrilateral. The vertices are the points A, B, C, and D. The sides are , and . The diagonals are and pCDA. and . The angles are pDAB, pABC, pBCD, Sides with a common endpoint are called adjacent or consecutive. Angles containing a common side are also called adjacent or consecutive. Two sides or angles that are not adjacent are called opposite. In the figure below, QABCD, QEFGH, and QIJLK are quadrilaterals; but not. Note that in naming quadrilaterals the order of the points is important, since quadrilateral. In these figures, would be a are diagonals, and are adjacent sides, while and are opposite sides. pJIL and pILK are adjacent angles, pEFG and pFGH are adjacent angles, pFEH and pFGH are opposite angles, and pABC and pADC are opposite angles. Notice that in the quadrilateral QABCD below: 1. 2. 3. 4. The diagonals intersect. Vertices are interior to their opposite angles. Sides lie entirely in one half-plane of their opposite sides. The diagonals lie between opposite vertices. Notice that in QEFGH, each of these is violated. We will call quadrilaterals like QABCD convex. But first, we need a theorem. Theorem: For a given quadrilateral, the following are equivalent: 1. Each side of the quadrilateral is on a halfplane of the opposite side of the quadrilateral. 2. The vertex of each angle of the quadrilateral is in the interior of its opposite angle. 3. The diagonals intersect each other (at interior points). 4. The diagonals lie between opposite vertices (i.e. opposite vertices are on opposite sides of the diagonal). G Proof: By the excited and clever students of Math 362. We take any of the equivalent statements in the previous theorem as the definition of a convex quadrilateral. Definition: If QABCD is a convex quadrilateral, then the angle sum is defined by Theorem: If QABCD is a convex quadrilateral, then . ~ Outline of proof: Given convex QABCD, we know that point C is interior to pBAD, so that . Similarly, because A is interior to pBCD, . Drawing the diagonal allows us to write the angle sum of QABCD as the sum of the angle sums of and . Since the angle sum for each triangle is less than or equal to 180, the result follows. Definition: A quadrilateral QABCD is a parallelogram if and . Theorem: Every parallelogram is a convex quadrilateral. ~ This follows immediately from version 1 of the definition of convex quadrilateral given above. For example, side of the line since otherwise this cannot happen since are exactly analogous. must lie entirely on one would intersect , and . The proofs for the other sides Theorem: Given a triangle , if A*D*B and A*E*C then QBCED is a convex quadrilateral. ~ First we must prove that QBCED is a quadrilateral. Note that segments do not intersect except at their common endpoints (this follows from the fact that A, B, and C are noncollinear, together with PSP and Incidence Postulate). This limits the ways in which the segments forming QBCED can intersect. intersect and or except at endpoints since they are subsets of . cannot intersect since it could only intersect at point A, and A*E*C. Moreover, line cannot at one point and so intersects analogous way we can show that can only intersect the only at D. In an exactly can only the “V Theorem” B and C are on the same side of cannot intersect at E. Finally, by and so . This shows that QBCED is a quadrilateral. To show that QBCED is convex, we note that A*D*B guarantees that D and B are on one side of , and A*E*C shows that E and C are together on one side of . Thus and are each contained wholly in one half-plane of their opposite sides. As noted above, the V Theorem establishes that B and C are on the same side of . Finally, the X Theorem (although it looks somewhat odd in this case) establishes that D and E are on the same side of . Theorem: If QABCD and QACBD are both quadrilaterals, then QABCD is not convex. If QABCD is a nonconvex quadrilateral, then QACBD is a quadrilateral. ~ Suppose QABCD and QACBD are both quadrilaterals. If QABCD is convex, then and , being diagonals, would have to intersect at an interior point. But that wold mean QACBD could not be a quadrilateral since two of its sides would intersect at a point other than an endpoint. To prove the second assertion, assume that QACBD is not a quadrilateral. Since adjacent sides can only intersect once at their common endpoints, the only way for QACBD not to be a quadrilateral would be if opposite sides intersected at interior points, i.e., if and intersected at interior points. But this would guarantee that was convex. We have proved the contrapositive of the assertion. Some interesting material not in the book: Definition: Two quadrilaterals QABCD and QXYZW are congruent under the correspondence ABCD : XYZW iff all pairs of corresponding sides and angles under the correspondence are congruent. In other words, this is another application of the basic definition abbreviated by CPCF. To no one’s surprise, we abbreviate this by QABCD QXYZW. Theorem SASAS Congruence): Suppose that two convex quadrilaterals QABCD and QXYZW are such that, under the correspondence ABCD : XYZW, three consecutive sides and the two angles included by those sides of QABCD are congruent, respectively, to the corresponding three consecutive sides and two included angles of QXYZW. Then QABCD QXYZW. G (Outline): Draw a pair of corresponding diagonals in the two quadrilaterals and use congruence criteria for triangles. Convexity allows us to add the angles when we need to (this is, in fact, a major reason why convexity is such an important idea for quadrilaterals), and the problem reduces to congruence for triangles. O Other Similarly Proved theorems: ASASA Theorem SASAA Theorem SASSS Theorem Question: What about ASAA? SSSS? In our work with quadrilaterals, it will be nice to use these theorems occasionally, rather than proving congruence by breaking things into triangles all the time.