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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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a2 and i want to talk about this statement and i’m going to refer to the

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

we talked about this statement in length but basically it means if you take any

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

already has a1 and it already has a3 and it already has the euclidean

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

think about it this line already has two points on it

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

lead to what we want and some of the cases might lead to contradictions if

they lead to contradictions that just means that case cannot happen

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

not happen and so what about this case right here though

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

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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

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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

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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