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## Geometry (all content)

### Course: Geometry (all content) > Unit 4

Lesson 8: Bringing it all together# Euler's line proof

Proving the somewhat mystical result that the circumcenter, centroid, and orthocenter all sit on the same line. Created by Sal Khan.

## Want to join the conversation?

- Is there any easy way of remembering which center is called what?

I keep forgetting...(15 votes)- No offense, but this is the scariest mnemonic device I've ever seen.(16 votes)

- Isn't the centroid of ABC also the centroid of DEF?(11 votes)
- It sure looks like it.(2 votes)

- So what about the incenter? Surely it must also have some fascinating properties?(5 votes)
- The incenter is the same distance from all the sides of the triange.(3 votes)

- Where can I use the Euler line in the world? And why is it important(3 votes)
- It is important to learn because a triangle is the basic shape in geometry, but it is filled with so many fascinating properties. It might not assist you in the world, but it is interesting to know.(7 votes)

- what is the 9 point circle? and does the incenter not lie on euler's line?(3 votes)
- In geometry, the nine-point circle is a circle that can be constructed for any given triangle. It is named this because it passes through nine points defined from the triangle. These nine points are:

The midpoint of each side of the triangle

The foot of each altitude

The midpoint of the line segment from each vertex of the triangle to the orthocenter (where the three altitudes meet; these line segments lie on their respective altitudes).

I hope this helps:)(4 votes)

- At9:11, Maybe you used that if vertical angles are same or alternate interior angles are same, vertices are on the same line. I can understand intuitively but can not understand logically..... Please explain more in detail.(4 votes)
- 7:12getting a little confused about how he can correlate the 2 distances... He can relate them with a ratio because the smaller medial triangle´s orthocenter is concurrent to the larger triangle´ circumcenter, right? So the perpendicular bisector of the larger triangle is also the height of the smaller medial triangle?(2 votes)
- At3:09Sal says that he needs to prove similarity between triangles FOG and CIG. At5:06after he uses alternate interior angles to show that angle F and angle C are congruent, couldn't he have used opposite angles to prove that angle OGF is congruent to angle IGC and conclude that all three angles on triangles OGF and IGC are congruent and therefore similar by the AAA apostulate? Thanks(1 vote)
- Are there any other purposes for the orthocenter, besides Euler's line and the 9-point circle?(1 vote)
- Couldn't you find that the two triangles are similar easier? I think I have a better way: By corresponding angles;<CGI = <FGO and <GCI = <GFO.

So, by using corresponding angles and AA similarity; Triangle FOG is similar to Triangle CIG.(1 vote)

## Video transcript

What I want to do in this video
is that for some triangle-- we're going to
focus on this larger triangle over
here, triangle ABC. What I want to do is prove
that the circumcenter of this triangle--
remember, the circumcenter is the intersection of its
perpendicular bisectors. That the circumcenter
for this triangle, the centroid of this
triangle-- the centroid is the intersection
of its medians-- and the orthocenter of
this triangle-- that's the intersection
of its altitudes-- all sit on the same line. Or that OI right over here
really is a line segment, or that OG and GI are really
just two segments that make up this larger line segment, 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, which
is the medial triangle for ABC. And there's already
a bunch of things that we know about
medial triangles, and we've proven this
in previous videos. One thing we know is
that the medial triangle DEF is going to be
similar to the larger triangle, the triangle it
is a medial triangle of. And that ratio from the
larger triangle to the smaller triangle is a 2 to
1 ratio, and this is going to be really
important to our proof. When two triangles are similar
with a given ratio, that means that if you take the
distance between any two 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 a medial triangle, and the triangle it is
the medial triangle of-- is that we've shown
that the orthocenter of the medial triangle is the
circumcenter of the larger triangle. So one way to think
about it-- point O, we already mentioned,
is the circumcenter of the larger triangle. It is also the orthocenter
of the smaller triangle. And we actually wrote up here. So point O, notice it
is on this perpendicular bisector over here. And I actually draw
a bunch of other ones in this dark gray
color, but I didn't want to make this
diagram too messy. But this is the circumcenter
of the larger triangle, and it is also the
orthocenter of the smaller triangle-- of DEF. And we actually
used this fact when we wanted to prove that
orthocenters are concurrent. We started with the
medial triangle. We said, OK, let's think about
where the altitudes intersect. And we said, well, look,
all of the altitudes are actually perpendicular
bisectors for our larger triangle if we
assume that this is the medial triangle of
that larger triangle. So point O, and this is going
to be important to our proof. It is a circumcenter
of triangle ABC, but it's the orthocenter
of DEF, and we've already talked about this
in previous videos. Now, in order to prove
that O, G, and I all sit on the same line--
or the same segment, in this case, what
I'm going to do is prove that triangle
FOG is similar is similar to triangle CIG. Because if I can prove that,
then their corresponding angles are going to be equivalent. You could say that
this angle is going to be equal to this
angle over here. And so OI would have to be
a transversal because we're going to see that these two
lines over here are parallel. Or if they have these two
triangles are similar-- so remember, we're looking at
this triangle right over here and this triangle over there. If they really are
similar, then this angle is going to be
equal to that angle, so these really would
be vertical angles. And so this really
would be a line. So let's go to the actual proof. So maybe I-- well, I
won't leave those two highlighted right there. So one thing, and I
hinted at this already. We know that this line
right over here-- we can call this line XC. We know this is
perpendicular to line AB. It is an altitude. And we also know that
FY right over here is perpendicular to AB. It is a perpendicular bisector. So they both form the same
angle with a transversal. You could view AB
as a transversal. So they must to be parallel. So we know that FY
is parallel to XC. Segment FY is
parallel to segment XC and we could write it like this. 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 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
angles are congruent. So that angle is going to
be congruent to that angle. So we could say angle OFG
is congruent to angle ICG. Now, the other thing
we know-- and this is a property of medians--
is that the centroid splits the median
into two segments that have a ratio of 2 to 1. Or another way to think
about is the centroid is 2/3 along the median. So we've proven this
in a previous video. We know that CG is
equal to 2 times GF. And I think you see
where we're going here. We have an angle. I showed 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
that the ratio of this side, CI to FO, is 2 to 1, then we
have two corresponding sides where the ratio is 2 to 1. And we have the angle
in between is congruent. We could use SAS similarity
to show that these two triangles are actually similar. So let's actually
think about that. CI is the distance between
the larger triangle's point C and its orthocenter. Right? I is the orthocenter
of the larger triangle. Well, what is FO? Well, F is the
corresponding point to point C on the
medial triangle, and we made sure
that we specified the similarity with the right--
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 its orthocenter. So this is the same
corresponding distance on the larger triangle and
on the medial triangle. And we already know they're
similar with a ratio of 2 to 1. And so, the
corresponding distances between any two points
on the two triangles are going to have
the same ratio. So because of that
similarity, we know that CI is going to
be equal to 2 times FO. I want to emphasize this. C is a 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 to
the orthocenter of the smaller triangle. The triangles are similar
in a ratio of 2 to 1. So the ratio of this
length to this length is going to be 2 to 1. So we've shown that the ratio
of this side to this side is 2 to 1. We've shown that the ratio
of this side to this side is also 2 to 1. And we've shown that the angle
in between them is congruent. So we have proven
by SAS similarity-- let me scroll down a little bit. By SAS similarity-- not
congruency, similarity. We've proven that triangle FOG
is similar to triangle CIG. And so we know corresponding
angles are congruent. We know that angle CIG
corresponds to angle FOG, so those are going
to be congruent. And we also know
that angle CGI-- let me do this in a new color. Angle CGI corresponds
to angle OGF, so they are also
going to be congruent. So you can look at
this in different ways. If this angle and this
angle are the same, you could now view
OI as a true line-- as a transversal of
these two parallel lines. So that lets you
know it's one line. Or you could look at
these two over here. Say, look, these two
angles are equivalent. So these must be
vertical angles. And so this must actually
be the same line. The angle that is approaching
this median right over here is the same angle
that it's leaving. So these are definitely
on the same line. So it's a very simple
proof, once again, for a very profound idea-- that
the orthocenter, the centroid, and the median of
any triangle all sit on this magical
Euler's Line.