Geometry (all content)
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.
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- 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)
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.