Proof: Triangle altitudes are concurrent (orthocenter)

What I want to do in this video is to show that if we start with any arbitrary triangle-- and this will be the arbitrary triangle that we're starting with-- that we can always make this the medial triangle of a larger triangle. And when we say the medial triangle, we mean that each of the vertices of this triangle will be the midpoint of the sides of a larger triangle. And I wanted to show that you can always construct that. If you start with this triangle, you can always have this be the medial triangle of a larger triangle. So to do that, let's draw a line that goes through this point right over here, but that's parallel to this line down here. So this line and this line up here are going to be parallel. So just like that. And immediately we can start to say some interesting things about the angles. So if we have a transversal right over here, we could view this side as a transversal of these two parallel lines, or of this line in the segment. We know that alternate interior angles are congruent. So that angle is going to be congruent to that angle. And we also know that this angle in blue, is going to be congruent to that angle right over there. Now, let's do that for the other two sides. So let's create a line that is parallel to this side of the triangle, but that goes through this point right over here. So let me draw it as well as possible. And so these two characters are going to be parallel, and you could always construct a line that's parallel to another line that goes to a point that's not on that line. And so once again, we can use alternate interior angles. We know that if this angle right over here-- let's say we have this orange angle-- it's alternate interior angle is this angle right over there. We also have corresponding angles. This blue angle corresponds to this angle right over here. So it will correspond to that angle right over there. And now let's draw another line that is parallel to this line right over here, but it goes through this vertex. It goes through the vertex that's opposite that line. And so let me just draw it. And you can always construct these parallel lines just like that. And let's see what happens. So once again, these two lines are parallel. So you could view this green line as a transversal. If this green line is a transversal, this corresponding angle is this angle right over here. If we view this green line as a transversal of both of these pink lines, then this angle corresponds to this angle right over here. If we view this yellow line as a transversal of both of these pink lines-- actually, let's look at it this way. View the pink line as a transversal of these two yellow lines, then we know that this angle corresponds to this angle right over here. And if you view this yellow line as a transversal of these two pink lines, then this angle corresponds to this angle right over here. And then the last thing we need to think about is if we think about the two green parallel lines and you view this yellow line as a transversal, then this corresponding angle in orange is right over here. This corresponds to that angle, because this yellow line is a transversal on both of these green lines. So what I've just shown starting with this inner triangle right over here is that if I construct these parallel lines in this way, that I now have four triangles if I include the original one, and they're all going to be similar to each other. And we know that they're all similar because they all have the exact same angles. You just need two angles to prove similarity. But all four of these triangles have the exact three angles. Now, the other thing we can show is that they're congruent. So all of these four are similar. And we also know they're congruent. For example, this side right over here in yellow is the side in this triangle, between the orange and the green side, is the side between the orange and the green side on this triangle right over here. So these two-- we have an angle, a side, and an angle. Angle-side-angle congruency. So these two are going to be congruent to each other. Then over here, on this inner triangle, our original triangle, the side that's between the orange and the blue side is going to be congruent to the side between the orange and the blue side on that triangle. Once again, we have angle-side-angle congruency. So this is congruent to this, which is congruent to that. All of these are going to be congruent. And by the same exact argument, this middle triangle is going to be congruent to this bottom triangle. You have an angle, blue angle, purple side, green angle. Blue angle, purple side, green angle. They're congruent to each other. So you have all of these triangles are congruent to each other. So their corresponding sides are equal. So if you look at this triangle over here, we know that the side between the blue angle and the green angle is going to be equal to this angle right over here. Sorry, equal to this length. So it's going to be equal to this length. Between the blue and the green we have this length, between the blue and the green we have that length, between the blue and the green we have that length right over there. So you immediately see that this point-- and let me label it now, maybe I should've labeled it before. If we call that point A, we see that A is the midpoint of-- let's call this point B, and call this point C right over here. So A is the midpoint of BC. So that's fair enough. So I was able to construct it in that way. Now let's look at the other sides. So this green side on all the triangles is the side between the blue and the orange angle. So between the blue and the orange angle, you have the green side, between the blue and the orange angle you have the green side. So once again, this length is equal to this length. And so if we call this point over here D, and maybe this point over here E, you see that D is the midpoint of BE. And then finally, the yellow side is between the green and the orange. So between the green and the orange, we have a yellow side. Between the green and the orange you have a yellow side. All of these triangles are congruent. So once again, let me call this F. We see that F is the midpoint of EC. So we've done what we wanted to do. We've shown that if you start with any arbitrary triangle, triangle ADF, we can construct a triangle BCE so that ADF is triangle BCE's medial triangle. And all that means is that the vertices of ADF sit on the midpoints of BCE. So you might say Sal, that by itself is interesting, but what's the whole point of this? The whole point of this is actually, I wanted to use this fact that if you give me any triangle, I can make it the medial triangle of the larger one to prove that the altitudes of this triangle are concurrent. And to see that, let me first draw the altitudes. So an altitude from vertex A looks like this. It starts at the vertex, goes to the opposite side, and is perpendicular to the opposite side. If I draw an altitude from vertex D, it would look like this. And if I draw an altitude from vertex F, it will look like this. And what I did, this whole set up of this video is to show, to prove that these will always be concurrent. And you might say, wait how do we know that they are concurrent? Well all you have to do is think about how they interact with the larger triangle. What are these altitudes to the larger triangle? Well, this yellow altitude to the larger triangle. Remember, these two yellow lines, line AD and line CE are parallel. So if this is a 90-degree angle, so its alternate interior angle is also going to be 90 degrees. So this right over here is perpendicular to CE, and it bisects CE, because we know that ADE is the medial triangle. This is the midpoint. So this right over here is perpendicular bisector. This is a perpendicular bisector for the larger triangle, for triangle BCE. So this altitude for the smaller one is a perpendicular bisector for the larger one. We can do that for all of them. If this angle right over here is 90 degrees, then this angle right over there is going to be 90 degrees, because this line is parallel to this, this is a transversal, alternate interior angles are the same. So this line right over here, this altitude of the smaller triangle, it bisects right at the midpoint of the larger one, on this side, and it's also a perpendicular bisector. So it's a perpendicular bisector of the larger triangle. And then finally, the same thing is true of this altitude right over here. It bisects this side of the larger triangle at a 90-degree angle. We know that because these two magenta lines the way we constructed the larger triangle, they're going to be parallel. So once again, this is a perpendicular bisector. So this whole reason, if you just give me any triangle, I can take its altitudes and I know that its altitude are going to intersect in one point. They're going to be concurrent. Because for any triangle, I can make it the medial triangle of a larger one, and then it's altitudes will be the perpendicular bisector for the larger triangle. And we already know that the perpendicular bisectors for any triangle are concurrent. They do intersect in exactly one point.