# The Affine Plane – What It Is and How It Works

(D4M) — Here is the video transcript for this video.

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hi everyone welcome back in this episode
the alphine plane what it is and how it works we’re going to describe what this
is and we’re going to get started uh with some theorems in this uh alphone plane
um now before i begin this episode is part of the series incidence geometry
tutorials with step-by-step proofs check out the link below in the description
so let’s do some math so we’re going to begin with the
question what is an alpha plane and in order to answer this question
we’re going to relate it to what we’ve been doing so far
so we’ve been talking about incidents and we’ve been talking about these three
axioms here axiom a1 every two points determines a unique line
and for every line there exists at least two points on it and axiom a3
there exists three non-collinear points and we did lots of episodes where we

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took these uh three axioms here and we proved uh some theorems and we worked on
some models um and so this is yeah this is the
incidence plane so now what i’m going to do is we’re going to look at what an
alphine plane is and so we’re going to have three
axioms here for an alpha plane and i wanted to put these three axioms here so
you can compare them next to each other so the first axiom for an alpha plane
is actually the same axiom for an instant plane uh every two distinct points are
incident with a unique line so um you know this is
pretty much the same thing it’s just here i used p’s and q’s here i just set
it out all right but here’s the difference here
between uh the incident plane and an alpha plane uh so remember in an
incident plane we we talked about these three axioms but then we considered
three additional statements um in the episode for example on parallelism we

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talked about the elliptic parallel property and the euclidean parallel
property and the hyperbolic parallel property and we gave models of these
three axioms and we gave different models where each
of those statements were holding so now we’re actually going to choose one of
those three we’re going to choose the euclidean parallel property and we’re
going to add this as an axiom so we’re going to keep a1 the same
and we’re going to switch out a2 and we’re going to put in here the euclidean
parallel property and then for our third axiom we’re going
to keep a3 exactly the same so the difference between an incident
plane and an alpha plane is this a second axiom here here we’re going to
put the euclidean uh parallel property right here
as our second axiom now i’m going to abbreviate as a pa um
pa basically is an abbreviation for playfair’s axiom
sometimes this is called playfair’s axiom so i’m going to refer to this is

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euclidean parallel property now i’m going to i’m gonna refer to it as pa
and the reason why i don’t use an epp is because
well the elliptic parallel property also goes by epp sometimes so to distinguish
them i’m just gonna call this a statement here pa and it’s basically
just the euclidean parallel property and again we did a previous episodes where
line l and you take any point p not on l then there exists a unique line passing
through p or incident with p that is parallel to l
there’s a unique line so that’s um exists exactly one line right so that’s the
um pa statement there playfair’s axiom or said you know euclidean parallel
property all right and so there’s the main
difference between these two right here now when we worked with the incident

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plane we started with these three axioms we went out and proved all these
theorems so i remember in the first couple of episodes we
proved each one of these statements in column format and we went through them
and every every you know every single step and then we
started getting better at them and then i asked you to provide column
proofs for these statements here and i hope that you did that but we also
wrote paragraph proofs um for them uh for you to see so check out those episodes
now i have a surprise for you this is really good i’d love this
so in an alphine plane so we have axioms a1 and a3 and we have
the occluding parallel property holding but you actually also get the incidence
axioms holding in other words if you replace axiom a2
and you put in the pa axiom well then you still get axiom a2 for free
that’s according to this theorem here and we’re going to prove this theorem

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here now so remember axiom a2 says uh for every line for
every line has at least two points on it every line has at least
two points on it and so that’s axiom a2 so what i’m
saying is that if you take axiom a1 and pa in a3
then you can now prove that a2 has to hold so that’s what this theorem here says
so let’s go through this and try to understand this and write up a proof
here so it remains to prove that a2 that there exists exactly
two distinct points on every line uh not exactly but there remains at least two
distinct points on every line so um we don’t need to reprove axiom a1

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because axiom a1 is already uh for alpha and plane so so so to be in it to show
the incident axioms hold technically we need to show axiom a1 a2 and a3 but this
parallel property in it this already has all the
so all that all that this is missing is a2 right so it remains to prove a2
and now in order to prove a2 remember we got to take any line and we got to show
there’s two points on it somewhere on it and so we’re going to start off with l
as a line it’s just an arbitrary line and somehow
using the other axioms and we can’t use theorem 1 through 11 yet because
remember these theorems that we proved these theorems assume that we have
already had axiom a1 a2 and a3 holding we assume we’re in an incidence geometry
so we cannot use these theorems here uh all we can use at this point here

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is axiom for alpha plane which is axiom a1 a3
and the euclidean parallel property so okay so let’s go and see this proof here
now so we’ve got a line l i’m going to try to keep a diagram running over here
we’ve got a line l okay now we can use axiom a3 because remember
alpine plane has axiom a3 holding so there exists three distinct points a b
and c that are not incident with the same line so a b and c cannot be on line l
now two of them might be or one of them might be or none of them might be but we
know that there are three non-linear points so all three of them can’t be on
line l but so somewhere we got these three points here um
so i’ll just put them over here but like i said we really don’t know
maybe one or even two or online l so let’s look at some of these cases here

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so um by axiom a1 so we have a line going through a and b and c
we have a line going through a b and we have a line going through ac
and we have a line going through bc so we have those three lines by axiom a1
so now we have four lines potentially we have these three lines here and we have
line l now it may happen that one of these lines is line l but if you
so if this is actually line l then we’re done that’s that’s there’s nothing else
to do because remember we’re trying to come up with two points on line l and if
this is equal to line l then a and b are on line l
if this one is equal to line l then again we’re done because all we need are
two points on line l and if this is line l then again we’re
done because we’ve got two points on line l
so basically we’re going to say if l is one of these lines then l is incident

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with two distinct points so we really want to rule out that case because in
that in those cases we’re actually done hence we assume that l is not any of
those lines because if l is any one of those three lines then the proof is over
we’re done we got two points on line l that that case is really you know done
all right so now let’s uh if l is incident with any two of these three points
right if l goes to a b or if it goes through ac
that’s basically another way of saying what we just said but i just wanted to
rephrase it by just talking about the points if l is
incident with any two of these three points then we’re finished we’re trying
to come up with two points on l so if l goes through any of these two points
we’re done that case is over all right so now we’ve got two cases
remaining here and i want to name them case one and case two
so assume l is incident with one of these points say point a that’ll be case one
and then case two will be assume that l is not incident with any of these three

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points here so once we finish these two cases here we’ll be done with the proof
because you know that’s all the possibilities we
talked about l passing through all three can’t happen we talked about l passing
through two of the three points we’re done
so what if l passes through just one of the points which we’ll just pick out
pick on a case one is really uh like three cases all rolled into one
um because if you say say point b well then you’ll be able to follow
through exactly what i’m doing but you’ll just switch out the a and the b
so we can just say without loss of generality say point a and then case 2 here
it doesn’t pass through any of the points there
okay so we got to do these two cases here we’re going to do case one first
and so let’s get started on that so here’s case one and let’s just put it
back up here assume l is incident with at most one of these three points again

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we’ve already talked about the case for three and two and now we’re going to
talk about l being incident with one of these points
so now let’s put the diagram back so we’ve got a line l
and we’re trying to come up with two points on it
and the case that we’re in is that we have uh three non-collinear points a b
and c and at this point we’re under the case where point a is um on line l
and so we have a line going through here um a b
and a is right here on this line and c is some point not on the line there
who knows where c is at except for the fact that we know that c is not on a b
line because then they would be nonco linear and we know c is not on line l
because we’re under the case where l is incident with only one of the three and
we’re picking on point a there so c is not on line l either all right so um
uh since a b and c are non-collinear b is not incident

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with the line through ac here um so actually i’m going to try to make
more room for my diagram here in the hopes of not um
getting in the way of the words over here so let’s say i have here
a line l coming through here and i’m going to put the c up here and the b here
and we got line l coming through here and we got a line
through those two points there okay so l is incident with a
and so remember the diagram is just a visual aid it’s just to help us read
the words here all right so since a b and c are
non-collinear p is not incident with this line going through a and c p is off
the line here all right and so what we can do with that so by
the euclidean parallel property or by plate ferrous axiom or i just say for

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abbreviation by pa there exists a unique line going through b
that’s parallel to ac i’m going to move b b down a little bit here so there’s a
unique line going through here let’s call that line m
and i drew them like they intersected over here but we don’t know that yet
that that’s the good thing about geometry is you have to be really
careful is how do you know you even have another point over here right so
but by pa right there’s a line that’s parallel to ac and it passes through b
in fact that line is unique all right so there exists a unique line
m incident with b that is parallel to ac here so remember that
m and this line here ac are parallel they have no points in common
all right so here we go we’re going to break case one up into sub cases
so case one is ellis incident with a that’s that’s case one but now i’m going

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to make some further assumptions but i’m gonna run through some cases here and
and then we’ll summarize and make sure that we got all the cases that we need
to get so my first case is i’m going to assume that l and m are parallel
and i’m going to assume that l and the line through bc are also parallel
now that may seem awkward to you um that those might actually hold but
remember when we’re trying to do cases we have to try to do all the cases
possible so we don’t have any gaps in our proof so some of the cases might
but in our proof we need to make sure consider all cases
so let’s consider this case here that l and m are parallel
and that l is also parallel to to the line through bc
so what can we say if if both of those if l if l is parallel to both of those

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well since uh both m and b c both of these lines right here since both of
these lines paths pass through b right the line through bc and the m they
both pass through b and they’re both parallel to l
so pa says there’s only one line that does that
so b is not on l right so that means there’s one line that passes through b
exactly one line that passes through b that’s parallel to l
so if both of them are parallel to l then in fact they have to be equal by
the by place axiom so these two lines have to be equal
but that poses a problem you see because c is on this line and
you know if they’re equal then that means c lies on m
but remember l a was the only point that’s on m uh sorry a was the only
point that’s on l that’s that’s the case that we’re in right up here so

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you know c lies on m but you know m is parallel to line ac so
this cannot happen so this case actually cannot happen these two lines both of
them cannot be parallel to line l and the main idea is by playfair’s axiom
tells us then the lines will be equal and well they can’t be equal and i
showed one way here why they can’t be equal but i also said another way
so in any case case 1.1 cannot happen this these two things cannot happen so
what would be another case all right so case 1.2 would be
well let’s suppose this right here doesn’t happen right
so what happens if l is not parallel to m what if they’re not parallel
well if two lines are not parallel right so parallel means they have a they do
not have a point in common right so this means it’s not the case that they don’t
have a point in common in other words they do have a point in common so l m

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have a point in common say point d which i’ll put right here
right or maybe i’ll make it look straighter
even though the word straight doesn’t have any meaning uh
so here we go point d right here um so l m intersect at some point and
i’m calling that point d now i drew it up here and again the diagram is not the
proof because according to the diagram there’s no way that point d and a can be
the same point but in our proof we actually have to go through that case
and we can’t rely upon the diagram oh look at the diagram a and d can’t be the
same no can’t do that right so if a and d are equal to each other then we’ll get
some problem hopefully and then we’ll have
two points on line l and we’ll be done right so let’s see if we get a problem
right here so let’s assume that a and d are actually equal to each other so
those are the same point now uh then by axiom a1

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these two lines have to be equal to each other so why is that so
um the line through a and b right here have points a and b on it
we know point b is on this line um because that’s how we came up with uh
the line b it was through through the point
that’s how we came up with the line m it was through the point b
so a and b are on this line right here b is on this line right here and if a is
equal to d so d is on d is on line m and if a is equal to d
then a is on this line also so that would put these two lines equal to each
other all right however these two lines cannot be equal to each
other because remember that m is parallel to this line here ac which
means they have no points in common a is
on this line so a cannot be on this line but if we claim that they’re equal
then a has to be on this line so this is a contradiction here so we cannot have

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this which means we cannot have a is equal to d all right so a is not equal to d
and so now l has two distinct points on it our line l um
the case assumption gave us point a and then
in this case we came up with the second point and in this case we said it could
so case 1.3 is what happens if l is parallel to line bc
so then we’re going to say that they’re not parallel so they have to have a
point in common and we’re going to call that point d again
so i’ll just draw this diagram here now so now we’re assuming l and line bc
are not parallel so here’s bc so let’s just put bc here and line l
coming up through here and it hits at a point d here now all right so line bc

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and line l are not parallel that’s our assumption for case 1.3 and
so l and l and line through bc have a point common call it d
um notice that a and d cannot be the same point again if a and d are the same
point then a is up here and a a and b and c are all all in the same line
and a b and c are non-collinear so this cannot happen
um so a cannot be equal to d again and so now again l has two distinct
points on it we don’t care that d is on this line right here um our ultimate
goal is just to get through all the cases and say either they cannot happen
or l has two distinct points on it if it if it possibly could happen
so now that we’ve gone through all of the cases right so we either have we
showed that the and cannot happen which means the negation happens and

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remember how you negate an and right and so we would negate this and have an or
and negate this and so we’ve gone through those cases we’ve gone we’ve
seen what happens if case 1.2 holds or if case 1.3 holds and in both of those
cases we got two points on l so cannot happen two points on l two
points on l all cases considered for case one so case one is over and l has
two distinct points on it all right so now let’s go on to case two
now oh by the way if you have any questions about case one or you know any
any questions at all put a comment below
and uh you know yeah i’d love to to hear your comments
all right so now let’s go with case two now so let’s get another diagram here
going so we got a line l and we have three distinct points and now uh
sorry uh three non-clean your points and
now we’re going to assume that a b and c are not incident with with the line at

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all right so we got let’s say here let me put the a b and c down here first so a
and then b and then c and now line l is just going to miss them completely
so i’ll put a and then b up here and then c right here and line now just
misses them completely that’s the case two here that we’re in so remember if l
goes through all three we said that can’t happen because they’re collinear
they’re not going here and if l goes through any two of the three we’re done
because we just need two points on l and we just did a case where l went through
one of the points and we got it so now this is our last case what if l misses
all three of these points how are we going to get two points on line l
well here we go uh we’re going to break case 2 up into sub cases
all right and so the first sub case will be
suppose that l and the line through b and c

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so let’s put the line through b and c here suppose those lines are parallel
so let’s see what happens in this case right here um now i’m going to make an
hypothesis here um so so this is our case hypothesis
and this is our case hypothesis here and so now i’m going to say if l is also
parallel to through the line through a b so if l is if if l is also parallel to
this line then you know remember we were we have
the euclidean parallel posture or p4 so p4 says that if you have a point
which we have b and it’s not on line l then there’s one unique line going
through b that is parallel to l so if this if if
bc is parallel to l and a b is parallel to l then they have to actually be the

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same line but they cannot be the same line because
then they would be collinear right so however this case cannot happen because
a b and c are non collinear so cannot happen that both of these lines are
parallel to l if both of these lines are parallel to l they have to be the same
line by p a and they cannot be the same line
right there so that’s a very useful uh observation right there so um
so these two points so these two lines right here have to be incident with the
point so let me just erase this now and say here we have a b and c and
we have c here and this line right here uh is not um so if if l is parallel here
then it cannot happen so this has to go through some point p here

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which i’ll just draw it right there right so so we got one point here on uh
line l so far but look look we had to assume all this we assume this this and
this um uh so okay so so that cannot happen
because then they would be all all this then they would be colleen here so it
has to happen that they’re not parallel which means they that we got some point
p here so now let’s do that case again but let’s do it with uh ac this time
so instead of a b now let’s use ac so again
if l is parallel to bc and l is parallel to ac
then we will get these two lines equal by pa and then
we’ll see that this case cannot happen these these two lines
um and actually that should be um equals right there

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so you know just like i had right here it should be equals right here so
if you’re watching these videos here you know that
i like to fix these things here as soon as i see them um
so where in the case here ac bc um i think that should be equals there
my typo here see if we can fix that real quick all right there we go
all right so if l is uh parallel to it through the line through ac
and we’re still assuming in case 2.1 that l is parallel to bc then those two
lines will have to be equal to each other by p4 there’s a unique line going
through b that’s parallel to l so if both of these lines are parallel to l
these lines have to be equal to each other however this is not the case again

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simply because a b and c cannot be on the same line because they’re not linear
all right so uh this line right here ac and l are incident with the point
and let’s call this point here q now again by the diagram by the by the
way the diagram looks it looks like p and q are not the same point
but again it has to be something that you’re that you have to prove
so first thing i want to say is that p both p and q are not a
so and the reason why is simple because p and q are on l
but a is not on l remember that l doesn’t hit any of these three points
so p and q are on l and a is not so they’re not equal to a um now let’s see here
what happens if p and q are equal well then these two lines are equal
which these two lines cannot be equal right here right so

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the line through uh a and c has p on it uh sorry it has q on it and
q is not the same thing as a here right so i got two points on this line and i
got those same two points on this line right here um a b
and so that will put these two lines equal to each other and we can’t have
that because they’re non-collinear right there
um so l is incident with two distinct points so we cannot have p is equal to q
here we have to have two they have to be different points here
and so there we go we have that l is uh not equal to
l has two distinct points on it now um we’re going to do the case here 2.1 like
we did and we’re going to do it again but instead of line bc here
the case for 2.2 will be for the case where l is parallel to line through a
and c and there’s nothing special about the b here

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so when we do the case 2.2 for the line ac here we’ll just replace uh this right
here instead of a b we’ll have an a and so we’ll just replace the b and a
everywhere here and we’ll have the exact
same arguments here so you don’t want to run through all of these arguments here
unless you want to do that on your own but the case for
the line ac here and the case for the line a b here you’ll get these two
arguments here and then you then you can argue that the two points you come up
with will be uh different from each other so uh these two cases right here
2.2 and 2.3 are entirely similar okay so the next case will be
2.4 will be now suppose l is not parallel to any of these lines right so
we’re going to have not parallel and we’ll have not parallel and we’ll have
not parallel and that’ll be the remaining case that we need to think
about here so if l is not parallel to any of those lines well then

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we have that line a b is incident with l at a point p
and we have line ac is incident at point q and you know we’ll just use the same
argument that we used before that p is not equal to a q
and so again l is incident with two distinct points
so this is a very clever argument it broke it down into cases right here
to get to the heart of the problem or the heart of the argument and then uh
argued on the cases right here all right and so that ends the proof
there so hope you like the proof hope you’re able to follow along
um sometimes when you first see a proof like this
you’ll want to rewrite it so that it makes sense for you
so the proof that i saw when i started reading this actually i saw
several of them but none of them i really liked none of them are really

00:31
followed very very easily so i rewrote the proof
and my proof was much longer than theirs but i wanted to show
a lot more steps for you but if this is not enough steps for you
then you should definitely write your own proof that’s longer moreover
because of the previous episodes we went through you might consider writing this
in column format although i think that such a proof would be
several lines long at least 100 probably but in any case there’s our
theorem here now keep in mind that now that this theorem has been proven now
that we know that in the alpha plane not
only do you have axiom a1 and a3 and you
have the euclidean parallel property but now we have axiom a2 also holds it
doesn’t hold as an axiom now now it holds as a theorem
but once action a2 has been proven now we get all of these theorems back so

00:32
all of these theorems actually hold in alphine plain geometry also so
this holds for instance geometry but it also holds for alpha geometry
but you know the alpha geometry has got this
pa axiom and one of the things you can do with the pa axiom is also show that
the axiom a2 holds but the pa axiom can actually we can prove a lot more with it
besides just these 11 theorems here now i want to refresh your memory in a
previous episode over parallelism we showed that if you have the axioms
a1a2a3 and you have the euclidean parallel posture holding which we do in
the alpha plane then we show that parallelism is an equivalence relation
so we did that in the previous episode so i’m not going to do that um again
um go back and check out that episode on parallelism where we went through this
theorem right here in great detail all right and so now i want to do uh one

00:33
more theorem um and this is a nice uh theorem here
um it has to do with parallel lines and i don’t think that you can prove this in
with just axioms a1 8283 i think we’re going to need the pa to prove this right
here so this is an example of the theorem that holds an alpha plane that
doesn’t hold in the incidence plane and let’s see what it is it says here um
in enough i’m playing if the line is incident
with and distinct from uh two distinct parallel lines so let’s say we have a
line l and m here so let’s say l and m and then we have a line that comes
through here it’s incident with one of them so let’s say here in line in
is incident with one of them then what this theorem here is saying is that then
it must pass through and become incident with the other one right here so if
these two lines are parallel eminem or parallel and they’re distinct lines

00:34
and one line comes through and in in this incident then it must follow
through and be incident with the other line as well so let’s see why this has
to hold so let’s start off with three lines here
okay and we’re going to assume that l and m are parallel
and we’re going to assume that n is incident with m
uh line n is incident with m all right so let’s just switch these around real
quick to make our diagram match our proof over
here all right so we’re assuming that they’re parallel three distinct lines
and in it hits the l a line m right there it passes through line m right
there all right so now by theorem one there’s uh is one and only one point and
remember we can use theorem one now because even though we’re in an alpha
plane the last proof that we just did showed that axiom a2 holds which means
we get those 11 theorems back and one of those theorems was

00:35
if you have two lines intersecting like this they’re going to intersect at a
unique point let’s call that point p with both m and n and so
we got point p here um now um if l and n are parallel
these two lines here are parallel we don’t want them to be we want them to uh
be incident but if they’re parallel so let’s see why that can’t happen
so so let’s just suppose it does happen so if they’re parallel then by p4
so if l and m are parallel here then by p4 these two lines have to be equal to
each other so why is that well this point p is not on line l
and if they’re both parallel to line l then they have to be the same line
by p four okay um hence um that cannot happen right because

00:36
these are these are not the same line we got three distinct lines so so if you
suppose this happens you run into that it cannot happen right
so now we know that it cannot happen and so n must meet line l right they’re
not parallel they’re not parallel they must meet somewhere
and that’s exactly what we were trying to show so that’s it right there that’s
not a long proof there it basically follows by p4 all right
that’s it hey let’s do some more math let’s do some more uh alpine playing
and i’ll see you in that episode