There Exist Three Distinct Lines That Are Nonconcurrent

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 in this episode we’re going to talk about
um this theorem right here in incidence geometry there exists three distinct
lines that are non-concurrent so let’s do some math
so let’s recap the incident axioms so we’re going to let point line and
incidence be undefined terms by the way this is just a recap so if this is going
too quickly for you you’re not following along make sure and check out the full
series the previous episodes the link is below in the description
so um axiom one which i just call a1 so for every point p and every point
uh q not equal to p right so if i take just uh sorry just a point and
i take another point q then there’s exist a new unique line
through those uh two points there so that’s incident axiom one

00:01
it’s in axiom two for every line l so now if i just put down a random line
any line you know just pick one there exist at least two points on that line
so i’ll just say p and q there’s just two distinct points
and then the third axiom is if uh so there exists three distinct points
uh with the property uh that no line is incident with all three of them so i
have a point p i have a point q and i have a point r
and i’m just randomly naming them the names are unimportant but i know
that there’s three of them and one line can’t go through all of them in other
words they’re non-collinear and so those are
that’s that’s one of the definitions that we have
three or more points are called collinear if there exists a line that
passes through all of them whereas incident with all of them
and then we have the dual concept would you switch points and lines three or
more lines are concurrent if there’s just a point incident with all of them

00:02
in other words the lines are concurrent if um they all
if a point if there’s at least one point incident with all of them
okay and lines l and m are parallel if they’re distinct right so they’re not
the same line and no point is incident with both of them right so here’s a line
l here’s a line m and they’re parallel meaning
they’re distinct and there’s no point on either one of them
all right so there’s just a really uh brief recap
and so what’s been proven in our geometry so far in the last episode we
proved this theorem here if we have two distinct lines that are not parallel
then there’s a unique point in common so l and m
um are different lines there’s a you know a point not on both of them
every point on l is on point m and every point on m is on l then the lines would
be equal but there are distinct lines and they

00:03
have a unique point in common only one point in common and so we proved that in
the previous episode there all right and so in today’s episode
we’re going to show that there exists three distinct lines that
are non-concurrent um and then in the upcoming episodes
we’re going to prove theorem three every
point is incident with at least one line
so in other words there’s no points that are just hanging out in no man’s land
um every point goes through every every point is on at least one line
and every line every line has at least one point that’s
not on it so if i just pick any line i know there’s some points that are not on
it so in other words all of our points aren’t aren’t on
a single line um and so number five every point has at least one line
not incident with it so if i just pick a random point i’m going to be able to
find a line that’s you know doesn’t pass through it

00:04
um and then theorem uh six um every point is incident with at least two
distinct lines so these are upcoming uh episodes uh
recommend checking them out in the series um and so yeah let’s get started on uh
theorem two there’s just three distinct lines that are non-concurrent
so we’re going to begin here um by making statements and justifications
and i’m going to number the statements so we’re going to make this in the
column format and then at the end we’ll do this again in
the paragraph format where the paragraph format the idea behind it is just to
contain the major important steps so that you can go back and reproduce
this proof which is the formal proof uh on your own
um and also i just want to say before we get started on this this isn’t
necessarily the shortest proof that you could write
this proof is just to help beginners get started writing proofs um and it’s just

00:05
for clarity purposes um you know my feeling is that any proof
that’s you know smaller than say a hundred lines
um isn’t a long proof and doesn’t need to necessarily be made any shorter
um unless it’s just completely obvious that it should be made shorter
but if a line is over that if it becomes too long then obviously you want to
fight hard to make it shorter and more digestible more readable
so anyways this proof is not necessarily going to be the shortest one
but it’s going to be try to be very clear
all right so we’re going to start off by naming these points
so by axiom three there exists three distinct uh nonconlinear points and i’m
just gonna call them a b and c now i’m not establishing any meaning to
the labeling so i just want to be clear just i know there’s three of them and so
i’ll name them a b and c and so by distinct here mean i know that

00:06
a is not b and a is not c and b is not c
and c is not a or b all right and so and they’re non-linear okay so we have that
by axiom three and now by axiom one see because we know that a and b are not
the same points so i have two distinct points here
and so action one says there’s a line that goes through them
i guess i could uh keep a diagram going over here so i have three nonconlinear
points by the way this over here is not just uh is not for
uh actual part of the proof it’s just to help with the words it’s just a
visualization aid so there’s just a line through a and b so i’ll put a line here
and there exists a line um through b and c and i’m not really naming the lines
other than just using this notation here this is
um the line because there’s a unique line so
this right here is just the unique line that goes through the points b and c

00:07
and then as you might imagine there’s a line that goes through a and c
also by axiom one so again you know you could probably put
all that in one statement uh and it’s just appealing to axiom one
but again i just you know want to make it you know nice clear and slow and easy
proof so now what i’m going to do is i’m going to make an r a hypothesis
and i’m going to say that a and b these two lines are actually equal to each
other so we know these three lines exist but when i say three lines i don’t
necessarily mean distinct so for example i might have here one
one and two and i wrote down three numbers but i only wrote down two distinct
numbers right so you know we want to be extra extra clear and very precise
so i want to say okay i have this line i have this line have
this line and i don’t know if they’re going to be equal to each other
using the notation that i wrote down it looks like they’re different but keep in

00:08
mind that that’s just notation so i want to say hey what’s so wrong
with them being equal what if they’re what if they are equal to each other so
that’s my my ra hypothesis which i want to come up with a contradiction now
so uh you know because i have a and b and c if if these lines are equal what
that means is that every point on this line is also on this line and every
point on this line is also on this line so step five right here tells us that
we have three collinear points now because a and b are on this line by step two
and b and c are on this line by step three
and step five says well they’re all on the same line so they’re all collinear
but that’s a contradiction because step one clearly says that they’re
non-collinear and step six says they are
collinear and so you know if you look at
the definition with uh clean cleaning or non-clean here they’re the you know

00:09
negation so um we have a contradiction there so in fact now we can conclude
that these are two different lines here now why did i pick on a and b uh sorry
why did i pick on this line through a and b and this line through b and c
because actually i you know could have done another one well we’re going to do
the other ones too what happens if the line through a and b
is equal to the line through a through a and c
well we’re going to get the same case again we’re going to get then that all
three lines are collinear and then we’re going to get that that’s
a contradiction because we know these three points are non-collinear
so now i know that these two lines are different
so and when you when we look at the diagram over here it’s it’s obvious our
intuition is oh yeah the lines are different right
but we’re actually going through the process what happens if ac is equal to
the line bc well then they’re all collinear they’ll
pass through the same line this line right here
and so then we get a contradiction again
so by looking at all three possibilities there

00:10
we know that we have three different lines here
so these are three distinct lines so there’s a good example of something
that you know by just looking at diagrams
you might have jumped to the conclusion from um you know step you know one
um you know a b and c are three distinct and uncle linear points and you might
have jumped to step 17 but if you wanted to see
how to actually fill in those steps well there’s some there’s some ideas on how
to do that okay so the next step is so there exists
a point incident with all three lines um and this is going to be our a
hypothesis this is exactly what we don’t
want to happen we don’t want there to be any point that’s on all three lines
because that’s what non-concurrent means so we’re assuming that it’s not
non-concurrent and that’s what this statement right
here means it’s not non-concurrent and we’re going to get a contradiction which

00:11
means it will be non-concurrent all right so i got this point x by assumption
all right so one and one uh must hold and i’m just gonna pick on the a we
could you could pick on the b or the c but i’m just gonna pick on the a so
because i have this point x and i have these other three points and i’m
wondering if x is any one of these three points or if x is not equal to the a
so that’s the law of excluded middle so this point x here
which is is incident with all three lines it’s somewhere on all three lines
and the first thing i’m going to do is say okay what if it’s a in other words
what if a is on all three lines here that’s case one
well then if a is all incident with all three lines then in fact they’re
collinear and we already saw that that can’t happen
so in fact that’s a contradiction by steps 1 and 22 they’re not collinear
now so i know that x is not equal to a so that’s case 2.
so this point here that’s on all three lines it cannot be a

00:12
all right so here we go um lines a b and a c are not parallel
and you can you know kind of get the into intuition from the diagram over
here the reason why they’re not parallels of course because point a is
on both of them um and and what we’ve uh sorry that should say
theorem one here by theorem one what we proved in the uh previous episode is
that if two lines are not parallel they have a unique point in common
and that’s this point here a is on both of them we can see that from right here
in the previous steps that a was on this one and a was on this one
um and then now because we’re under the assumption that a is on all three lines
it’s definitely on these two lines here also
and so here’s this point a that’s on these two non-parallel lines and x is on
these two non-parallel line so the uniqueness of theorem one says that in

00:13
fact those two points have to be the same but we’ve already shown that x is not
equal to a we’ve already tried to do that and we got a contradiction so
there does not exist a point in other words this um
hypothesis the r a hypothesis that we had in step [Music] 18
um cannot um you know cannot happen so there’s not a point incident with all
three lines right so that’s the negation of step 18 which was our our
which was the raa hypothesis all right very good so these lines are
not concurrent and that’s why the definition of non-concurrent and that’s
our conclusion is that we have three non-concurrent lines so um
you know just to make sure i didn’t lose anybody when we got to the x here
um we’re assuming here that step 18 is there exists a point somewhere in all

00:14
three lines and my argument was was if x is a well we get a contradiction
if x is not a well then you know because a is on both of these
two lines and so is a then in fact x has to be a and then
but that leads us back to the same contradiction that we had before
so x you know it can’t exist um and so they’re uh non-concurrent
all right and so now let’s look at a paragraph proof
um something you know because this took us 29 lines now again it could be
written it could have been written up shorter that’s not really the point the
point is though is what were the big main ideas of the proof
certainly using axiom a3 was an important part of the way that i
organized this proof and then the uh you know there’s three points here and then
another uh big point of this proof was we got these lines through these two
points here through these three points and then another part was we use theorem

00:15
one also that was theorem 1 what used right here so
let me see here if i can actually go and uh theorem one here
all right so let’s see if i can fix that um
error right there real quick there we go all right so the point is is that those
are the three important steps that you know the
proof should have here so let’s go through a paragraph proof here now so
this is how you might write it up um and in in my opinion um the proof that i’m
going to write for you is to give you the main ideas so that
you can go back on your own without much difficulty at all
and write the line-by-line proof so i’m going to say hey i’m going to use
axiom a3 i got three justine to not clean your points so that’s important

00:16
because you know if you have say a hundred axioms which we don’t or if you
have a hundred theorems to sort through and realize you know what theorems do
you need you know that could be very important part of your proof is by
saying what am i using all right and then by axiom a1 and i
applied it three times we got this we got these lines here here i’m not saying
three distinct lines because we don’t know that right now at this point right
here so if any one of these lines is equal to the other
then all three points are collinear so this sentence right here kind of
summarizes several of the steps that we’ve laid out line by line by line in
the column proof and because you know not everyone wants
to read the 29 lines because some people are able to prove this on their own and
all they need are these ideas hence these three lines are distinct
so you know that just summarizes the steps that we did all right um
hence these three lines are distinct and then what did we have after they were

00:17
distinct so we have here uh the remainder of the proof here
where we’re going to go here and say step 18 was so
hence all three lines are distinct so now we have three distinct lines
and now we can say we don’t want to end the proof yet so
scratch that right there the end of the proof here what we’re going to say is
that assume for a contradiction so this is another way of saying raa
hypothesis and conclusion assume for a contradiction there exists a point
and i’m going to call it x there exists a point x incident with all three lines
all right and so the three lines the only three lines i mentioned are these

00:18
three right here so i’ll just say with all three lines here
all right and so then we can say if x equals a then a is incident
then if x equals a then a b and c are collinear are collinear and this would be
collinear and then i will say um are collinear contrary to hypothesis
in other words that’s a dead end because you know we’re assuming that
where they’re not linear if x is not equal to a then um [Music] lines a b

00:19
and uh and then then distinct lines then distinct lines a b and a c um
do not have a unique point in common do not have a unique point in common
contrary to theorem one so if x is not equal to a right so we got a b and and
and uh ac so we have these two lines here and if x is not equal to a um
but it’s on all three lines so you know then then we’ll have uh we won’t have a
unique point in common because x is not equal to a
so um you know we’ve already proved here one is true right so um

00:20
therefore or hence so hence um x does not exist and so what that says is that
so therefore the these lines here therefore a b a c and bc
are non-concurrent just by definition all right and so there’s the uh all the
main ideas and that proof right there and and you know your paragraph proof
can be shorter than that um it just kind of you know needs to
summarize all the steps in the proof uh the more drafts you write of it the
the quicker it will get and you’ll get the hang of it keep writing drafts don’t
settle on your first draft of your proof and as we work through this uh series
our proofs will get shorter and shorter as you become more capable improving

00:21
them on your own all right so that’s it for this episode
right here uh check out the uh next theorem and the next episode right here
we’re gonna prove um some more interesting theorems in our
incidence geometry and i’ll see you then

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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