Quadrilateral in which diagonal is partially outside?

Question

I just want an example of a quadrilateral in which a diagonal lies "partially outside" the quadrilateral. By that, I mean that some part of a diagonal must be outside and some part inside the quadrilateral.

Obviously the quadrilateral will be concave iff one diagonal lies completely outside the figure, but now, I saw a problem related to concave quadrilateral, so to be sure, I looked it up and in the definition, it said

Concave quadrilaterals are four sided polygons that have one interior angle greater than $180^{\circ}$. We can identify concave quadrilaterals by using the fact that one of its diagonals lie partially or completely outside the quadrilateral.

So, I tried to come up with a concave quad with diagonal partially outside, like equilateral triangle with centroid and so forth, but none worked,

So,Tldr; I just need one example a (concave) quadrilateral in which diagonal lies partially outside.

Answer 1

Two vertices are adjacent to each diagonal of a quadrilateral.

If the remaining two vertices are on opposite sides of the diagonal, then the diagonal is entirely inside the quadrilateral.

Otherwise if the remaining two vertices are on the same side of the diagonal, then the diagonal is outside the quadrilateral. (Assuming the quadrilateral is still simple and not self-intersecting)

Either way, the diagonal is entirely in or entirely out.

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From one vertex adjacent to a diagonal, there are two paths along the quadrilateral sides to the opposite vertex. Each path passes through one intermediate vertex.

With just one intermediate vertex, each side-vertex-side path lies entirely on one side of the diagonal.

So the diagonal does not cross any side of the quadrilateral. (Which is necessary for the diagonal to be partially inside and partially outside the quadrilateral)

Answer 2

We may prove that a quadrilateral with a partially external diagonal does not exist via a pigeonhole argument.

A quadrilateral and its diagonals define six lines in the plane. In nondegenerate cases such a set of lines would have $15$ total intersections, including those at infinity where lines are parallel. But with the sides and diagonals of a quadrilateral three lines are concurrent at each vertex, and each such concurrency reduces three otherwise distinct points of intersection to one. So the number of distinct intersection points is reduced to $15-(4×2)=7$. Out of those seven, four are the vertices, two are the intersections of opposite pairs of sides and one is the intersection of the diagonals. Thus no opportunity for any other intersection such as a diagonal crossing a side.

The text may have used the phrase "partially or completely" for consistency with other polygons. Given five or more sides, it's easy to design examples of polygons with partially external diagonals; the numbers in what would be a pigeonhole-type exclusion do not add up the same way as we found for quadrilaterals.