Euler's Line Proof

Proving the somewhat mystical result that the circumcenter, centroid and orthocenter

What I wanna do on this video, for some triangle we're gonna focused

on this larger triangle over heretriangle ABC

What I wanna do is prove that circumcenter,

remember the circumcenter is the intersection

of it's perpendicular bisectors

The circumcenter for this triangle, the centroid of this triangle,

the centroid is the intersection of it's medians,

and the orthocenter, that's the intersection of it's altitudes

all sit on the same line, or that OI right over here really a line segment

Or that OG and GI are really just 2 segments

that make up these 2 line segments which is part of the euler line

And to do that, I've set up a medial triangle right over here,

triangle FED, or actually I should say, triangle DEF,

whch is the medial triangle for ABC

And there's already a bunch of things

that we know about medial triangle

and we've proven this in previous videos

One thing we know, is the medial triangle DEF,

DEF is going to be similar to the larger triangle

the triangle that is a medial triangle of,

it is of, and ratio from the larger triangle to the smaller triangle

it's a 2 to 1 ratio

And this is really important to proof

When two triangles are similar with the given ratio,

that means if you take the distance between any 2 corresponding parts

of the two similar triangles that ratio will be 2 to 1

Now the other relationship that we've already shown,

the other relationship between the medial triangle,

and the triangle that is the medial triangle

of is that we've shown the orthocenter

of the medial center of the larger triangle

So one way to think about it is Point O,

we already mentioned is the circumcenter of the larger triangle

It is also the, it is also the orthocenter of the smaller triangle,

we actually wrote it of here

so Point O noticed it is on this perpendicular bisector,

you know, I should do a bunch of other ones of this dark grey color

but I didn't wanna make this diagram too messy

But this is the circumcenter of the larger triangle

Now in order to prove that OG and I all sit on the same line

or the same segment in this case

What I'm going to do, I wanna prove,

I'm going to prove that triangle FOG,

I'm going to prove that triangle FOG,

is similar to, is similar to triangle CIG,

is similar to triangle CIG

Because if I can prove that, then their corresponding angles

are going to be equivalent,

you could say this is angle is going to be equal

to those angle over here

And so OI would have to be a transversal,

cause we're goiong to see these two lines here are parallel

or if these two triangles are similar,

just remember that someone's look at our triangle here

and this triangles over there

If they truly are similar then this angle

is going to be equal to that angle

which would mean that,

so these are really would be vertical angles

and so this really would be real line

So let's go the actual proof

So maybe, I don't need those to highlight it over here

So one thing and I've hinted about this already,

we know that this line over here, we can call this like XC

we know this is perpendicular like AB

it is an altitude, and we also know that FY right over here

is perpendicular to AB, it is a perpendicular bi-sector

So they both form the same angle with a transversal

you can view AB as a transversal

so they must be parallel, so we know that FY,

FY is parallel to XC, to XC

segment FY is parallel to XC,

segment FY is parallel to segment XC

And we can write it; this guy is parallel to that guy there

And that's useful because we know that alternate interior angles

of a transversal, when a transversal intersects two parallel

lines are congruent

So we know, we know that this angle, so we know that FC is a line

it is a median of this larger triangle, triangle ABC

So you have a line intersecting two parallel lines,

alternate interior angels are congruent

So that angle is gonna be congruent to that angle

So you could say angle OFG is congruent to angle,

so it's OFG, it's congruent to angel ICG,

to ICG, now the other, the other thing we know,

this is a property, this is a property of medians is that

a median splits up or should I say the centroid,

splits the median in to two segments that have a ratio of two to one

or another way to think about this is a centroid

is two thirds along the median

So we know we've proven this on a previous video

We know that CG, CG is a equal to 2 times GF, 2 times GF

and I think you see where you are going,

we have an angle, I've shown you that the ratio of this side

to this side is 2 to 1 and that's just the property

of centroids and medians

And now if we can show you the ratio of his side CI is FO is 2 to 1

that we have two corresponding sides where the ratio is 2 to 1,

and we have the angle and between this congruent,

and we have the SAS singularity to show that these two triangles

are actually similar, so let's actually think about that

CI is the distance between, CI is the between the larger triangle's

point C orthocenter of the larger triangle

Well what is FO?

Well F is a corresponding point to point C on the medial triangle

and we make sure that we specify the similarity with the right, F,

F corresponds to point C

So FO is the distance between F on the smaller medial triangle

and the smaller medial triangle's orthocenter

So this is the distance between C

and the orthocenter of the larger triangle

This is the distance between the corresponding

side of the medial triangle

and the smaller medial triangle and it's orthocenter

So this is the same corresponding distance

on the larger triangle and the medial triangle,

and we already know that they're similar with the ratio of 2 to 1

And so the corresponding distances between any 2 points

on the two same triangle are gonna have the same ratio

So because of that similarity, because of the similarity

we know that CI, CI is gonna be equal to 2 times FO

I wanna emphasize this; C is the corresponding point to F,

when we look at both of these similar triangles,

I is the orthocenter of the larger triangle,

O is the orthocenter of the smaller triangle

You're taking a corresponding point

to the orthocenter of the larger triangle,

corresponding point of the smaller triangle

The triangles are similar to the ratios of 2 to 1

so the ratio of this length to this length is going to be 2 to 1

So we've shown, we've shown the ratio

of this side to this side is 2 to 1

We've shown the ratio if this side to side is also to 2 to 1

We've shown the angle in between are,

the angle between them is congruent

So we have proven by SAS similarity,

so it goes down a little bit, so by, by SAS similarity,

not congruency, similarity we've proven that triangle FOG

is similar to CIG

And so we know corresponding triangles are congruent,

we know that angle CIG correspond to angle FOG

so those are going to be congruent, and we also know that angle CGI,

angle CGI, let me do this a new color,

angle CGI corresponds to angel OGF

so they're also going to be congruent

So you can look at the different ways of these angle

and this angle are the same you can now view OI

as a true line as a transversal of these two parallel lines

So that let's you know that's a one line

or you can look these two over here,

so look these two angles are equivalent

so these must be vertical angles,

so this must actually be, this actually must be the same line

The angle that this is approaching, this,

this median is the same angle that is leaving

So these are all, these are definitely on the same line

So it's a very simple proof, once again,

from a very profound idea, the orthocenter,

the centroid and the median of any triangle

all sit on this magical euler's line