A quadrilateral $ABCD$ is formed from four distinct points (called the vertices), no three of which are collinear, and from the segments $AB$, $CB$, $CD$, and $DA$ (called the sides), which have no intersections except at those endpoints labeled by the same letter. The notation for this quadrilateral is not unique- e.g., quadrilateral $ABCD$ = quadrilateral $CBAD$.
Two vertices that are endpoints of a side are called adjacent; otherwise the two vertices are called opposite. The remaining pair of segments $AC$ and $BD$ formed from the four points are called diagonals of the quadrilateral; they may or may not intersect at some fifth point. If $X$, $Y$, $Z$ are the vertices of quadrilateral $ABCD$ such that $Y$ is adjacent to both $X$ and $Z$, then angle $XYZ$ is called an angle of the quadrilateral; if $W$ is the fourth vertex, then angle $XWZ$ and angle $XYZ$ are called opposite angles.
The quadrilaterals of main interest are the convex ones. By definition, they are the quadrilaterals such that each pair of opposite sides, e.g., $AB$ and $CD$, has the property that $CD$ is contained in one of the half-planes bounded by the line through $A$ and $B$, and $AB$ is contained in one of the half-planes bounded by the line through $C$ and $D$.
a) Using Pasch's theorem, prove that if one pair of opposite sides has this property, then so does the other pair of opposite sides.
b) Prove, using the crossbar theorem, that the following are equivalent:
- The quadrilateral is convex.
- Each vertex of the quadrilateral lies in the interior of the opposite angle.
- The diagonals of the quadrilateral meet.
I can’t seem to make sense of a) at all. For b) I approached it in this way – I made three separate proofs:
- If the quadrilateral is convex, then the diagonals of the quadrilateral meet.
Proof: Assume quadrilateral $ABCD$ is a convex quadrilateral. We have to prove that segment $AC$ and segment $BD$ have a point in common. By the
definition of a convex quadrilateral, $C$ is in the interior of angle
$DAB$. Hence, ray $AC$ intersects segment $BD$ at some point $E$ (by the
crossbar theorem). Therefore, $E$ is the required intersection point
of the diagonals $AC$ and $BD$.
- If the diagonals of the quadrilateral meet, then each vertex of the quadrilateral lies in the interior of the opposite angle.
Proof:
- If each vertex of the quadrilateral lies in the interior of the opposite angle, then the quadrilateral is convex.
Proof:
I’m also confused over the proofs for 2. And 3.. Theorems and axioms that might be helpful:
Pasch’s Theorem: If $A$, $B$, and $C$ are distinct points and $l$ is any line intersecting $AB$ in a point between $A$ and $B$, then $l$ also intersects either $AC$, or $BC$. If $C$ does not lie on $l$, then $l$ does not intersect both $AC$ and $BC$.
Interior an angle: Given an angle $\angle CAB$, define a point $D$ to be in the interior of $\angle CAB$ if $D$ is on the same side of Ray $AC$ as $B$ and if $D$ is also on the same side of Ray $AB$ as $C$. Thus, the interior of an angle is the intersection of two half-planes.
Crossbar Theorem: If ray $AD$ is between ray $AC$ and ray $AB$, then ray $AD$ intersects segment $BC$.
Any help at all would be much appreciated.
Best Answer
In a possible attempt to explain a), let us focus solely on a single angle, say angle $A$. Similarly, draw tangent lines extending from the two adjacent sides, namely $AB$ and $AD$. Assuming $A\neq180$ (which we can, because it would cause $ABCD$ to be a triangle), $AB$ and $AD$ are not parallel.
This means that they meet at $A$ and continue, getting further apart as they go. If $A \lt 180$, meaning $ABCD$ is convex, $AB$ and $AD$ continue away from the shape, not intersecting any sides.
However, if $A \gt 180$, $AB$ and $AD$ enter the interior or $ABCD$ after intersecting at $A$. As the lines are infinite and the quadrilateral is not, the lines must at some point leave the shape. As two lines can only meet at a single point, and will not intersect themselves, they must leave the shape through one of the other two sides (Note Pasch's Theorem).
As both $AB$ and $AD$ are equally dependent on the angle of $A$, it is not possible for only one of the two lines to split one of the other sides.