Inscribing convex polygon within simple polyon Suppose you are given a simple (but not necessarily convex) polygon $C$ and a point $p$ inside this polygon. I have a particular way of inscribing a convex polygon $I$ within $P$, and I would like to know (a) if this is a known concept from geometry and (b) if there exists an efficient algorithm for computing $I$.
The idea is the following:


*

*For each edge of $C$, find the closest point to $p$ on that edge.

*If the closest point is not an endpoint or is an endpoint that is not a reflex angle, extend the edge into a line.

*If the closest point is an endpoint $v$ that is a reflex angle, draw a line that passes through $v$ and is orthogonal to the line segment from $p$ to $v$.

*For each line from (2) and (3), construct the half-plane containing $p$ bounded by that line.

*$I$ is the polygon created by the intersection of all half-planes from (4).


$I$ will be convex because it is the intersection of half-planes.
For example, in the following figure, the solid black boundary forms $C$, the blue lines are the ones constructed in step (1) orthogonal to the corresponding dotted lines, and the gray region is $I$.
Below is the original construction, which @HSN pointed out was completely incorrect.
Suppose you are given a simple (but not necessarily convex) polygon $C$ and a point $p$ inside this polygon. I have a particular way of inscribing a convex polygon $I$ within $P$, and I would like to know (a) if this is a known concept from geometry and (b) if there exists an efficient algorithm for computing $I$.
The idea is the following:


*

*For each vertex $v$ of $C$ that forms a reflex interior angle (between 180 and 360 degrees), draw a line that passes through $v$ and is orthogonal to from $p$ to $v$.

*For each line from (1), construct the half-plane containing $p$ bounded by that line.

*$I$ is the polygon created by the intersection of $C$ with all half-planes from (3).


For example, in the following figure, the solid black boundary forms $C$, the blue lines are the ones constructed in step (1) orthogonal to the corresponding dotted lines, and the gray region is $I$.

 A: I'm not convinced this construction actually works for any $p$ and $C$ and I don't think I've come across this particular construction before. Why am I doubtful?
To make things somewhat easier, let's give all vertices in your picture a name. The point at the left-hand bottom is $A$, at the right-hand bottom $B$ and continuing counterclockwise up to $H$. The two vertices having a reflex interior angle are $E$ and $F$. We will elongate the line segment $EF$ and let $S$ be the intersection of this line with $AH$.
Now, look what happens if we choose $p$ inside of the rectangle $SFGH$. The perpendicular to $pE$ through $E$ cuts off a triangle having $ED$ as a side (and a third point on $CD$). When also using the perpendicular to $pF$ through $F$, we also get a trapezoid shape having $EF$ as a side.
I'm not good at quickly making pictures that look alright and don't like using a link that may expire, but the following is a horrible illustration of what I mean. Maybe someone would like to improve it. As you can see, the result consists of two polygons joined at the vertex $E$ and is not convex.

