... HOW?!

[size=150][size=200] How do we know that Euler didn't just make this up? [/size][/size]

[size=100][size=150][br][br]If you're anything like me, some math facts are just SO cool that you can't believe it. That's why we use mathematical proofs: to guarantee the truth of such a statement.[br][/size][/size][br]Want to understand Euler line a little more? This activity is lengthy, so it's not required that you read it through. Although, it is full of glorious mathematics, so I encourage you to at least try!

[b]Let's take a look at one of the proofs for Euler line. [/b]

We are given triangle XYZ and it's orthocenter O, circumcenter C, and centroid N.

We will have to make a few constructions to help us through the proof.

First we will start with the medial triangle of triangle XYZ. The medial triangle is the triangle formed by the midsegments of the larger triangle.

The medial triangle of triangle XYZ is triangle EFG, and is labeled in yellow. Something that we should recall, and will be important in this proof, is that [size=150][b][size=100][i]a triangle and its medial triangle are similar with a ratio of 2:1[/i][/size][/b][/size]. [br]Thus we can say that [math]\bigtriangleup[/math]XYZ[math]\sim\bigtriangleup[/math]EFG.

Next we look at constructing line XO, an altitude of triangle XYZ. Recall that a triangle's orthocenter is created by the intersection of its altitudes, therefore we can say that a line going through a vertex and the orthocenter is a altitude of triangle XYZ.

We're also going to construct line EC. Recall that the circumcenter is created by a triangle's perpendicular bisectors. Because this line goes through point C (the circumcenter) and point E (a midpoint), we can say that it is one of the perpendicular bisectors of triangle XYZ.

Lastly, we'll add line EX, which contains the median of triangle XYZ, segment EX. Recall that the centroid of a triangle is created by the intersection of its medians. Therefore, we can say that point N lies on line EX.

[i][color=#666666]cool constructions... where's the good stuff?[/color][/i]

Let's look at what information these constructions tell us.

Because segments EF, FG, and GE are midsegments, we can say that they are parallel to their corresponding sides of triangle XYZ. We then have the following:[br][br][b]YZ // FG[/b][br]ZX // GE [br]EF // XY[br][br]For the sake of this proof, we will only take the first one in bold along with us on our diagram.

We know that YZ is perpendicular to XO since XO is an altitude of triangle XYZ. [br]We also know that YZ is perpendicular to EC by the definition of a perpendicular bisector.

Since line XO and EC are both perpendicular to YZ, we can say that lines XO and EC are also parallel. [br][br]WELLLL since those are parallel, we can conclude some things about the angles they form with their transversal EX. [br][br]Recall that point N is on line EX. If we look at [math]\angle[/math]CEN and [math]\angle[/math]OXN, we will see that we have a set of alternate interior angles. By AIA theorem, we can then say that [math]\angle[/math]CEN[math]\cong\angle[/math]OXN. These angles are highlighted in purple.

We're going to take a quick detour, so hold that thought.

If we construct the [b][color=#a64d79]orthocenter[/color][/b] of the medial triangle EFG, you'll see in the figure below that point C, the [color=#6aa84f][b]circumcenter[/b][/color] of triangle XYZ, is the same point. Does that mean a triangle's [color=#6aa84f][b]circumcenter[/b][/color] is the same point as its medial triangle's [color=#a64d79][b]orthocenter[/b][/color]???????

Still with me? We're almost there!

Recall that [math]\bigtriangleup[/math]XYZ~[math]\bigtriangleup[/math]EFG. Since segment XO is the distance between a vertex of triangle XYZ and its orthocenter, we can say that the corresponding part of triangle EFG is proportional. That corresponding part is EC (the distance between triangle EFG's orthocenter and its vertex), therefore XO and EC are proportional!

Let's look at segment EX.

Recall that EX is a median of triangle XYZ, and the intersection of triangle XYZ's median create centroid, N. [br][br]What do you know about the distances between a triangle's centroid, vertices, and midpoints? [br][br][br]THINKKKKKK!

There's a ratio...

The distance from centroid to vertex, and centroid to the midpoint of the opposite side, is proportional by a 2:1 ratio!!! [br][br]That means that the distance from vertex X and centroid N, and the distance from midpoint E to centroid N, has a ratio of 2:1. Therefore, EN and XN are proportional.

Ready for more triangles?!

Let's pull out SAS similarity theorem. [br][br]First side: EC is proportional to XO[br]Second side: EN is proportional to XN[br]Included angle: [math]\angle[/math]CEN[math]\cong\angle[/math]OXN (remember that from way back there?)[br][br]THEREFORE: [br][br][math]\bigtriangleup[/math]CEN~[math]\bigtriangleup[/math]OXN by SAS~ theorem.

Did you forget what we're even trying to prove yet?

Here's a quick reminder... we want to show that circumcenter C, orthocenter O, and centroid N of triangle XYZ are collinear.

[b][size=150]What do we know about similar triangles and their corresponding angles? [/size][/b][br][br]Ahhhh yes, they are congruent!

Therefore we have that [math]\angle[/math]CNE[math]\cong\angle[/math]ONX. [br][br]Since point N is on line EX, the segment ON and line EX create a linear pair. [br][br]Remember what's special about a linear pair? Their angles are supplementary! If [math]\angle ONX+\angle ONE=180^\circ[/math] and [math]\angle CNE\cong\angle ONX[/math], then [math]\angle CNE+\angle ONE=180^\circ[/math] too. Angles that add to [math]180^\circ[/math] are supplementary, and therefore are a linear pair, which then implies points [color=#6aa84f][b]C[/b] (circumcenter)[/color], [color=#3d85c6][b]N[/b] (centroid)[/color], and [color=#c27ba0][b]O[/b] (orthocenter)[/color] are... COLLINEAR!!!! [br][br]And there you have it - [i][b][color=#ff0000][size=100][size=150]The Euler Line[/size][/size][/color][/b][/i]