Bringing it all together
Euler's Line Proof Proving the somewhat mystical result that the circumcenter, centroid and orthocenter
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- 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
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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