WEBVTT

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

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on this larger triangle over heretriangle ABC

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What I wanna do is prove that circumcenter,

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remember the circumcenter is the intersection

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of it's perpendicular bisectors

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

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the centroid is the intersection of it's medians,

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

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

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Or that OG and GI are really just 2 segments

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

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

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

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whch is the medial triangle for ABC

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And there's already a bunch of things

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that we know about medial triangle

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and we've proven this in previous videos

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One thing we know, is the medial triangle DEF,

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DEF is going to be similar to the larger triangle

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the triangle that is a medial triangle of,

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

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it's a 2 to 1 ratio

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And this is really important to proof

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When two triangles are similar with the given ratio,

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

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

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Now the other relationship that we've already shown,

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the other relationship between the medial triangle,

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and the triangle that is the medial triangle

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of is that we've shown the orthocenter

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of the medial center of the larger triangle

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So one way to think about it is Point O,

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we already mentioned is the circumcenter of the larger triangle

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

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we actually wrote it of here

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so Point O noticed it is on this perpendicular bisector,

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

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but I didn't wanna make this diagram too messy

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But this is the circumcenter of the larger triangle

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

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or the same segment in this case

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

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I'm going to prove that triangle FOG,

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I'm going to prove that triangle FOG,

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is similar to, is similar to triangle CIG,

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is similar to triangle CIG

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Because if I can prove that, then their corresponding angles

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are going to be equivalent,

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you could say this is angle is going to be equal

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to those angle over here

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And so OI would have to be a transversal,

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

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or if these two triangles are similar,

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just remember that someone's look at our triangle here

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and this triangles over there

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If they truly are similar then this angle

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is going to be equal to that angle

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which would mean that,

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so these are really would be vertical angles

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and so this really would be real line

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So let's go the actual proof

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

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

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

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we know this is perpendicular like AB

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

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

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So they both form the same angle with a transversal

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you can view AB as a transversal

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

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FY is parallel to XC, to XC

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segment FY is parallel to XC,

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segment FY is parallel to segment XC

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

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

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of a transversal, when a transversal intersects two parallel

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lines are congruent

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

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it is a median of this larger triangle, triangle ABC

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So you have a line intersecting two parallel lines,

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alternate interior angels are congruent

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So that angle is gonna be congruent to that angle

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So you could say angle OFG is congruent to angle,

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

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

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

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a median splits up or should I say the centroid,

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

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or another way to think about this is a centroid

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is two thirds along the median

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

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

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and I think you see where you are going,

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

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

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of centroids and medians

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

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

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and we have the angle and between this congruent,

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

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

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

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point C orthocenter of the larger triangle

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Well what is FO?

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

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

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F corresponds to point C

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

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and the smaller medial triangle's orthocenter

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So this is the distance between C

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and the orthocenter of the larger triangle

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This is the distance between the corresponding

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side of the medial triangle

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and the smaller medial triangle and it's orthocenter

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So this is the same corresponding distance

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on the larger triangle and the medial triangle,

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

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And so the corresponding distances between any 2 points

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on the two same triangle are gonna have the same ratio

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So because of that similarity, because of the similarity

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

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

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when we look at both of these similar triangles,

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I is the orthocenter of the larger triangle,

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O is the orthocenter of the smaller triangle

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You're taking a corresponding point

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to the orthocenter of the larger triangle,

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corresponding point of the smaller triangle

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The triangles are similar to the ratios of 2 to 1

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

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

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of this side to this side is 2 to 1

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

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We've shown the angle in between are,

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the angle between them is congruent

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So we have proven by SAS similarity,

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

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not congruency, similarity we've proven that triangle FOG

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is similar to CIG

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And so we know corresponding triangles are congruent,

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we know that angle CIG correspond to angle FOG

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

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angle CGI, let me do this a new color,

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angle CGI corresponds to angel OGF

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so they're also going to be congruent

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So you can look at the different ways of these angle

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and this angle are the same you can now view OI

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

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

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or you can look these two over here,

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so look these two angles are equivalent

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so these must be vertical angles,

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

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The angle that this is approaching, this,

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this median is the same angle that is leaving

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

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

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from a very profound idea, the orthocenter,

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the centroid and the median of any triangle

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all sit on this magical euler's line