As future educators at Normandale Community College, our Math 1060 class explored four commonly-used methods for calculating the sum of interior angles of polygons. These methods fell into two camps. We determined the number and sizes of the interior/exterior angle pairs, then subtracting the sum of the exterior angles. We also looked at three different ways of dividing a polygon up into triangles where all the triangles shared a common vertex.
The basic methods we used were proven to work for polygons with any number (n) sides/vertices. The only hitch was that, like most geometry lessons, our testing only used regular or other simple convex polygons.
We wanted to know (and by "wanted to know" I mean "were assigned the task of proving") whether these methods would also work for concave polygons. If a method fails, can it be adapted so that it would work?
For the purposes of this packet we will use octagons to test our ideas at first. If we can modify a method so it works for all octagons, it should able to be modified to for any polygon.
With that, let's get started, shall we? Please pull out your octagons. If you didn't bring one with you, why don't you just sketch one on a piece of paper...
First: a pet peeve of mine. You drew this didn't you?
What about this vaguely "fishy" shape? Isn't it an octagon too? eight sides, count em. I submit that it is just as much an octagon as your stop sign above, and we're going to use it as one.
Or this cool supervillian-logo looking thing, can't it get some love?
"Count to eight, don't discriminate."
OK, I've made my point... I'll get off my high-horse and back to work now. sorry.
Where were we? Oh yeah, Internal Angles sums...
The basic approach is to sum the interior/exterior (supplementary) angle pairs, and subtract 360 (the sum of eight 45-degree exterior angles.
Consider, The octagon consists of 8 "straight lines" (180 degrees each). Were you walking around it counter -clockwise, you would see (and presumably measure) the angle on your left as the "interior" of the octagon.
To your right, you'd see/measure the angles that were on the exterior. At each vertex, you "turn through" 45 degrees to face the new direction. At the end, you have done this 8 times. Since those external angles are not part of the interior of the octagon, they must be subtracted. This works for polygons of any number of sides... at least the convex ones.
So what happens when this formula meets a concave octagon? Let's take a look.
Yikes. Well, we do still have eight pairs of interior/exterior angles, but since some of them "dent inward" they will have internal angles greater than 180.
In this case, "sigma", (the interior angle under the fish's tail) is 247 degrees and the 67-degree angle "lambda" we would normally think of as external resides inside the polygon.
This brings the total for each dent-inward angle to 314 degrees. Shouldn't all these angles be supplementary? So you know that something isn't right about that. Yes. Yes, it seems... fishy. (sorry)
It no longer fits our formula, but let's check it. Calculated this way, the fish-octagon's internal angles total 1281. This is well over the 1080 we are expecting. Yeah... this is no good.
This problem can be corrected by adjusting how you calculate those dent-inward angles. If you treated the three 67-degree "lambda" angles as negative in your calculation, subtracting them from the 1281 total would get us back to 1080 degrees.
For a better explanation, see Christopher Danielson's Packet on this correction technique, by clicking here.
Let's look at the three formulas that divide up a polygon into triangles, and how they break down.
This seems easier to visualize for many. All interior angles (vertices) are accounted for once.
Each internal angle of the polygon is completely accounted for: either as a vertex of one triangle (like vertex B), or by having it's measure split into more than one.
For example, vertex E is divided in half, part if it is measured in triangle ADE, and the rest is accounted for in triangle AEF. Similarly, vertex A is accounted for with 22.5 degrees in each of the 6 triangles.
Once again, our trusty regular octagon does a fine job of showing us what we need to see. 
"Why is it (n-2) ?" you ask. Because there is no triangle formed by connecting vertex A to vertices adjacent to it (B or H). Each triangle needs three sides. Thus there are two fewer triangles than sides.
What happens when we throw the fish-octagon against this method? Let's see.
As you can see, this won't fit our formula for several reasons.
First, there are new triangles that exist outside the polygon (IBC, and JEF), and we don't know how or even whether to include those in our calculations.
Second, what do we do about the angles created at J and I? I see an additional 360 degrees when you add those two supplementary (straight line) angles.
Thirdly, and most glaringly, how on earth do we deal with the two quadrilaterals formed at the fish's tail? DEFH and DCBH are not even triangles!
The next method also divides the polygon into triangles, this time there are 7 of them, and our formula reflects that (n-1). We have to subtract 180 degrees (-180) to account for the angles centered at the point labeled "I". They are not "interior angles" of this regular octagon. They are just a product of the measuring process. Once we do that, we get the 1080 we expected to see. 
Like it did in the previous "divide-into-triangles example", this approach breaks down when we use it on a concave polygon. Again, we have extra triangles outside the polygon near the tail (JBC, FEK). We also see extra angles at J and K that would need to be accounted for, or subtracted.
This method is a lot like the previous two. You subdivide the polygon into triangles. Each of the polygon's vertices are accounted for in the resulting triangles. Knowing that every triangle contains 180 degrees, and remembering that you have to subtract 360 degrees from your total to account for the angles you created at I (once around the middle). Our trusty stop sign (n=8) looks something like this. 
and, not surprisingly we run into many of the same problems with this concave octagon. Triangles get formed outside the polygon, additional angles on the polygon that may/may not be included, etc.
I know. You're thinking to yourself... I can solve that! I'm a clever sort. I can just move the point labeled "I" to a place where it won't even cause those problems in the first place....
Ooooh. That really IS clever. Good for you. Good for all of us! No weird new triangles to adjudicate, no extra angles to measure. We can just plug in our formula! 180n - 360 = 1080. The day is saved and we're... oh, wait.
We're supposed to see if it works for EVERY one, not just the fish.
Remember this guy? There's literally no place inside this octagon you could put the "I" point without it being a big mess.
Leave it to a super-villian to spoil everything again.
Great. We have three ways to calculate the interior angle sums, and none of them are bulletproof. They can all be foiled by pushing just one side until an interior angle gets bigger than 180 degrees.
Fortunately, we needn't throw the baby out with the bathwater. I may be dating myself with that expression, but hear me out. Perhaps we can salvage these ideas.
I submit that the thing screwing up everything is that we are restricting ourselves when we force all the "dividing and measuring" triangles to share a common point. I think if you just follow a few simple guidelines, you can draw triangles however you like.
First: You have to include all the interior angles once. An angle can be split up (as we've seen before) but you can't leave any parts unaccounted for.
Second: It's most efficient to restrict your triangles only using vertices they have in common with the polygon. That way, there's no additional angles that need to be subtracted. You can use any vertices you like, in any order.
Third: If you do add a point for creating triangles, whether out of necessity, or for the sake of making the triangles look pretty (or for whatever reason) - you must account for them. You have to subtract 180 degrees for any points on the polygon's sides, or 360 degrees for a point created inside the polygon.
I think that should effectively include all interior angles, and exclude anything that is not an interior angle.
Let's look at a few to be sure.
Our old friend, the regular octagon. Looks pretty good. There are n-2 triangles with no additional angles or triangles to muddy things up, and all interior angles are completely used. But does this work for concave polygons?
Phew! the fish checks out. 
Even the super-villian logo is subdued.
What about something that is not an octagon (just in case)? Here's a decagon (10-sided)
(n-2) or 8 triangles times 180 degrees = 1440.
Even this irregular decagon works. It's still got (n-2) or 8 triangles. 8* 180 = 1440.
All right! The formula works just fine.
I think they should, let's take a peek. With I on the side of the polygon, we see (n-1) gives us 7 triangles. Multiply that by 180 = 1260 Subtract 180 and we are back at 1080 as expected. Nice.
Mixing it up:
What if we put a point on the inside AND on the edge of the polygon? It shouldn't matter... as long as you account for everything... and let n = the number of triangles created.
With "I" located inside the polygon as the "eye" (see what I did there?), AND including all sorts of line segments making triangles around the tail. We can still follow the formula as long as we pay attention to the guidelines on using subtracting anything we created and make sure to include the polygon vertices properly.
In this case, The figure has 12 triangles (12*180 = 2160)
But we've added 2 internal points, around which there are 360 degrees we need to subtract (I and L).
There are also two new side points on the top and bottom. (J and K, at 180 degrees each)
So how does this check out?
2160 (all the interior triangles' angle sums) - 360 ( I) - 360 (L) - 360 (J+K) = 1080.
There you have it! The formulas can still work, just remember to account for anything you have added to the mix when making your triangles in your formula.
Sure. On one side of the equation you want to have the sum of the triangles you wind up with. That's what all the "(n-2)", and "(n-1)" was for anyway, right?
(total triangles * 180) - (anything you introduced to make those triangles) = interior angle sum
Here's the going rate
point on the polygon's side = 180
interior intersection of two segments or point inside the polygon = 360
btw:
You could even use shapes other than triangles in your equation just substitute the correct degree amounts as needed.
quadrilateral = 360
pentagon = 540
etc.
But let's not bite off that chunk today. Perhaps another packet, another day.
