Incidence Axioms and Basic Theorems

Video Series: Incidence Geometry (Tutorials with Step-by-Step Proofs)

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

00:00
hi everyone welcome back i’m dave um we’re gonna go over the incident
axioms and some basic theorems that we can prove using these axioms
so uh this episode is part of the series instance geometry tutorials with
step-by-step proofs the link is below in the description let’s do some math
so let’s go over the instant axioms first we’re going to let point line and
incidence be undefined terms so we’re thinking about incidence as being a
relation here for example a point lies on a line
or we say point is incident with a line and we have three axioms axiom one is if
you’re given two distinct points p and q then there exists a line that passes
through them in its unique line and the second axiom says
if you take any line you can find two distinct points on you can find two
points on it and the third axiom says there exists three distinct points with

00:01
the property no line is incident with all three of them
and then we have our three basic definitions uh points are collinear if there is
exist a line that passes through all of them
lines are concurrent if they all have a point in common and in parallel
lines are called parallel if they have no points in common
so you can rewrite axiom 3 to say there exists three non-collinear points
which is a nice succinct way of looking at axiom three
in the previous episodes we have talked about uh so some these theorems here
we’ve proven theorem one and theorem two three four five and six
so we proved these six theorems so far so we’re going to prove some more
theorems in today’s episode um but unlike these other episodes we did
everything column format and then we wrote a paragraph proof
today we’re just going to write them in paragraph proof i’m going to write them

00:02
in such a way that i hope that you’ll be able to write your own column proof uh
for them and and i heartily recommend that you do that
so before i prove each of the new theorems not d6 we’ve already done these six
i recommend pausing the video and trying to write a column proof yourself
and then coming back and looking at what i did uh now before we go on though
i want to go over some notation and a couple of other things
in case you haven’t seen the previous episodes
so for points i’m going to use capital letters so for example a b c
there’s three points there and when i write a b and c i don’t necessarily mean
that they’re different points so if i say if i mean that they’re
different i’ll say three distinct points and just because i say there’s three
points here that doesn’t mean that they’re all distinct from each

00:03
other so just caveat there’s i’m very precise when i mean that they’re
different from each other um so if we um i’ll also use sometimes a p and
a q and an r for points also for lines i like to use lowercase
letters l m and n will be common notation for a line
um and then we have this right here so this is something that is really not
standard anywhere uh it’s just geometry is just
done by so many people it’s just hard to make a standard so
by axiom a1 if we’re given any two points say p and q
there’s only one line going through them or if i have say points a and b there’s
only one line going through them if they’re two distinct points
so um how the question is how to uh you what’s the notation for the actual
line right because we’re not using diagrams for notation right so i want to say

00:04
e any one of these three right here the notation for them
so this is the notation i’ve been using so far it represents um
the line passing through two distinct points a and b
so that that’s that’s what this right here is notation for
um the line the unique line redundant right the unique line
uh incident um incident with a with incident the unique line incident with
distinct points a and b so as you can see that this notation right here
uh represents all of these words and so this notation is very handy the unique
line incident with two distinct points a and b

00:05
and so i’ll use this right here also with a comma sometimes
and then sometimes without a comma the comma is really unnecessary
but some people like to have it but anyways all three of these are
notation and they just are representations of these words the
unique line incident with two distinct points a and b
um l is a unique line passing through two distinct points a and b
so uh that’s the other thing i wanted to talk about was
um you know there’s other ways of saying incidents
and so i’ve been i’ve been doing this but i just want to make it uh you know
pointed out so we can say uh p is incident or i’ll say um a is incident a is

00:06
incident with line l all right and so that’s the word the
official word is incident and i say also a lies on line l and sometimes
i won’t even say point point a lies on line l a lies on l um
or we can say l passes through a where line l passes through point a okay so i
mean the same thing for all of them they all mean the same things just a
different way of saying just intuitive ways of saying the word incident incident
so we want to be rigorous as much as possible but we also want to have
intuition as much as possible we want everything don’t we all right so

00:07
let’s go on now uh the fourth thing for today is going to be theorem seven like
i showed you a minute ago we’ve already proven theorem one through six we did
column format first in the previous episodes and we did the proofs and so
today right now i’m just going to work on the proofs here and i’m hoping that
you’ll pause the video right here and try to write out your own column proof
and i’ve shown you how to do that kind of six times so you know if you haven’t
seen those episodes go back and try them and try to write your column proof so
your column proof should have statement justification and you should number each
and every line and so what i’m going to do here though is just actually write up
a proof here all right so hopefully you paused already so here’s the proof here
so i’m going to use axiom a1 and it’s going to happen that these two lines are
going to be equal to each other now the reason why these two lines are going to
be equal to each other is because c and a are distinct so why are cna distinct

00:08
right so c is uh distinct from a right so it says right there c is
distinct from a um c and a are on this line for sure and
they’re distinct from each other axiom a1 says there’s only one line that goes
through cna because cna are just two distinct points
but cna are also on this line right a is on this line and our hypothesis is that
c is on this line so c and a also go through this line but
axiom one says there’s only one line that goes through this line so axiom one
tells us immediately that these two are equal because i know that c and a are
distinct and they’re incident with both lines so those lines have to be equal to
each other by a1 so there’s my first step
so by a1 i’m going to use the same logic but instead with c and a i’m going to
use c and b so by a1 so line through the line through b and c
is also equal to the line through a and b so why is that because c and b are

00:09
distinct right and it’s telling us if c is distinct from b right so so uh b and
c are obviously on this line but they’re also on this line b is on this line
right here but it says that c is on this line right here also right there so b
and c are on both of these lines so by a1 these lines have to be equal so when
i say by a1 right here i’m justifying this equal sign right here uh by by
saying this right here um and so that tells us that this line is equal to a b
and this line is equal to a b and so we have exactly what we want
all three lines have to be equal to each other
and so you know the intuition behind that is if c is on this line right here
so we have a line given by this two distinct points here a and b and it sees
somewhere on that line and it’s not a and it’s not b
so c could be here here here you know wherever it’s at but the line going

00:10
through c and a is equal to the line going through c and b
which is equal to the line going through a and b they’re all the same line as
long as c is somewhere on that line um yeah so uh let’s go to
um nick’s theorem here now theorem eight all right so again pause the video and
try it out and so here’s going to be our proof uh if
these two lines are given and b and c are distinct points
then the lines through a and b is equal to the line uh through b and c
so here’s our proof so as you might imagine again we’re going to use a1
to get some lines are equal to each other so the line through a and b is unique
and how do we know it’s equal to the line through b and c
well first off we know there’s a line through b and c by a one there’s a line
going through it and it’s unique so b and c are on this line

00:11
b is on this line right here um and we’re we’re assuming that these two
lines are equal which means where c is on this line so c is on this line
so in fact b and c are on both of these lines so by a1 they have to be equal
right so that’s how i say that here by a1 it follows these two lines are equal
there’s a unique line that goes through b and c
beings here on both of these lines here all right so then oops
so that’s the proof of uh theorem eight there it’s very very short there
it follows by by a1 right there all right let’s try another one um theorem 9
so if l is any line let’s just draw this here if l is any line
and let’s see here we’ve got a line so then there exist lines m and m

00:12
and they’re distinct m m m n are not the same
but they both uh have a point in common with l so you know like here’s an m and
here’s an n so i have to be able to find those lines
right there how can i find two lines that are incident with this arbitrary
line l all right so here’s the proof so i recommend pausing it and trying to
write it column proof again all right so first thing i’m going to do
is right so l is any line so i’m going to say l is a line
uh now we have a previous theorem remember we already proved theorems one
through six in the previous episode so the m4 says
if you take any line you can find some point not on that line
so let’s just keep a diagram going over here here’s a line l and by theorem 4
so there’s some point p right here not on that line l it’s important that it’s
not on that line l all right and so there exist points a and b incident with

00:13
l let me move my p out of the way that sounded funny um there exist points a and
b on line l okay there’s a and b uh why is there exist points a and b
uh incident with line l actually i forgot to
put that in here this is theorem nine i like to be as complete as possible so
while you’re thinking why that happens i’m going to fix this here okay so
all right sorry about that typo there uh or not typo but just uh oversight there
so by a2 axiom 82 remember a2 says no matter uh any line you have there’s two
points on it so i have this line here l and axiom a2 says there’s at least two

00:14
points on every line all right and so now you might imagine
how we’re going to come up with the lines m and n so um
now we we’re going to know that a is not
p why is a not p well p is not on line l a is on line l right so i i say right
here a is not equal to p but the way that you would write write that up in a
in a column format is you would say a equals p um r a a hypothesis
and then you would say oh well then that puts p on line l because because a is
but p is not on line l so you get a contradiction and then you would say a
is not equal to p and you would get an r r a a conclusion from that right so i’m
skipping those uh small little tiny steps there that i think that you can
try to fill in for yourself so a is not p and then b is not p also right if if p

00:15
was equal to b then p would be on line l but you know line l was made uh so i
mean p was made so that it was not online now okay so so these are not
equal to each other so what that means is by axiom a1 right there’s a line going
through a and p and there’s a line going through b and p
right now what we know is that these are distinct points a is not p and
b is not b and a and b are incident with line l and
p is not instant with line out so i tried to justify that a little bit there
so i didn’t skip any steps there all right so by a1 and i’ll move my
diagram down here again so we got line l we got p not on the line and we got our
a and b on the line here as as we know by axiom a1 there exists
lines through these two points here so there’s a line going through here which

00:16
is line m and this is point p right here and there exists a line going through p
and b which i’ll just put right there and this is line n right there
so by a1 we can we can draw a line through a and p
which i’m i’m just going to call it l that’s what i mean by that notation
right there just call it just call it m and there exists a line through b and p
and i’m just going to name it you know it exists by the axiom so i just give it
a name and both these lines have a point in common with l
so what is the point in common with l right so a is on line l and a is on this
point uh and a is on this line and b is on line l and b is on line m in
right so both of these lines have a point in common with l
now the question is are they distinct right and so then here i say clearly l m
and n are distinct lines so um you know why do i say that why can’t they be the

00:17
same well this is again where you would try some r a a hypothesis in your column
proof so you would say something like um you know they are equal all right so if
they’re equal right well uh p is on line m um and if they’re equal then p is um
um online n uh and there’s nothing wrong with that so let’s see here
if they’re all if both of these lines are equal to each other then
they’re all equal to each other because uh a is online m
and if they’re equal to each other that would put a on line in
and if they’re equal to each other that would put b on line m
and so l m and n would all be the same line and then p would be on that line
right so we cannot have these right here that would tell us that a and b are on

00:18
all three lines making all three lines equal to each other by
a one here okay so again i i suppose you can
fill in those small little steps there clearly they’re distinct lines if you
try to set them equal to each other you will
run into a contradiction let’s just say that all right and so let’s look at uh
theorem 10 now um again try to pause and see if you can get the proof there
so if a is any point so there exist points uh sorry a is any point
uh then there exist points b and c uh such that there’s no line going
through all three of them there’s no line going through all three of them
they’re non-clean year so let’s see how to prove that
so i’m going to start off with a is a point and now remember by theorem 5
there exists a line non-incident with a so there’s a line here which i’m just

00:19
going to call it l and we know by axiom a2 that every line has two distinct
points on it and i’m going to name them b and c
there exists distinct points b and c and i’m going to use these two points to
satisfy the theorem um and their incident with line l right
so every line has two points on it two distinct points
now what happens if we uh think that there’s one line going through all three
of them what’s so wrong with that so i’m
going to assume for a contradiction that
they are co-linear and i’m going to call
that line m so m is the line that passes through all three of them
well if it passes through all three of them it passes through b and c
but axiom a1 says there’s only one line that passes through b and c
so if l so if m is a line that passes through all three
it certainly passes through b and c and so it has to be equal to l

00:20
but a was a point not incident with l right so there’s no way that there can
be a line going through all three of them so this contradiction shows right
the contradiction is that a is incident with l and a is not incident with l
so that’s a clear contradiction so we cannot have a line going through all
three of them so in fact they cannot be collinear which means they are in fact
non-collinear all right so there’s a theorem 10 and we got one more theorem here
it’s a very similar theorem so this time we have two points so let’s
say we have point a and b and let’s call it a and let’s call it b
um then there exists a point c such that a b and c are non-linear
so i have a point c here there we go and they’re non-collinear
so how we come up with point c well pause the video now and give it a try

00:21
all right here we go let a and b be two distinct points
and so there exists a line um a through a b and
i didn’t say why there exists a line through a and b and i usually like to be
thorough so i’m gonna go back here just very quickly
and so that there there should be a bi theorem one here or axiom one right
okay got it all right here we go so by axiom a1 uh a1
is that a one or an exclamation mark here we go so by uh by a by axiom one by
a one there exists a line through a and b all right good we’re going to call

00:22
that line l and and by theorem four they’re just a point
not incident with this l so i’m going to call that point here c
so by theorem four we know that there’s some point not on the line there all
right so i’m going to assume for contradiction now that i got these three
points i’m going to assume that there’s a line that goes through all three of
them and i’m going to call that line l now we’re hoping that’s not true but we
got to see that it’s not true what if it is true all right so if i get a line
that goes through all three of them so by a1
um the lines are equal and why why well because a and b are two distinct points
on line l and m also goes through a and b so
there’s only one line a1 says there’s only one line that goes through a and b
so they have to be equal and so that means that c is on line l

00:23
but c was not on line l so this contradiction right here shows that this m
cannot exist in other words they cannot be collinear
so they must be non-collinear here we go
all right and there we go we have a kind of a summary right there of what we’ve
done so far um we did theorems one through six in
previous episodes and then in today’s episode we did seven eight nine ten and
eleven and i thought i would just put them all on one screen for us for us to
see there all right so uh what we got going on next is
parallelism yeah check out this episode right here to uh do some more math and
i’ll see you in that episode

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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