Download Quadrilaterals Definition: If A, B, C and D are any four points no

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.