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Proof: Triangle altitudes are concurrent (orthocenter)

Showing that any triangle can be the medial triangle for some larger triangle. Using this to show that the altitudes of a triangle are concurrent (at the orthocenter). Created by Sal Khan.
Video transcript
What I want to do in this video is to show that if we start with any arbitrary triangle, this would be the arbitrary triangle that we're starting with starting, that we can always make this the medial triangle of a larger triangle When we say the medial triangle we mean that each of the vertices of this tria, of this triangle will be the midpoint of the sides of a larger triangle I wanted to show you that you can always construct that If you start with this triangle, you can always have this be the medial triangle of 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 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 can view this side as a transversal of these two parallel lines or this line and this 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' create, 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, that goes thru 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 Just let's say we have this orange angle, its 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'll correspond to that angle right over there And now let's draw another line that is parallel to this line right over here, parallel to this one right over here but it goes through this vertex, Goes to the vertex that's opposite that line And so Let me just draw it You can always construct these parallel lines, and 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 transversal, this corresponding angle is this angle right over here If we view, 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, 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 would have to, the last this we should, we need to think about is, if we think about the two green parallel lines, the two green parallel lines, and you look at the and you view this yellow line as a transversal Then this corresponding angle in orange is right over here This corresponds to that angle Cause this yellow line is a transversal on both of these green lines So what I've just show 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, that 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, the, 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 sides 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, same exact argument this middle triangle is going to be congruent to this bottom triangle That the, you have an angle, blue angle, purple side, green angle, blue angle, purple side, green angle, they're congruent to each other So, if all of these triangles are congruent to each other So the corresponding sides are equal So if you look at this triangle over here We know that the side between the blue angle, between the blue angle and the green angle is, is going to be equal to this angle right over here, and sorry, it is 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 have labeled it before If we call that point A, we see that A is the midpoint of, of, we 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, is the, 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 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, triangle ADF, we can construct a triangle BCE, we can construct, construct a triangle BCE, so that ADF, ADF is triangle BCE's medial triangle And all that means, 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 is I wanted to use this fact that if you give me any triangle, I can make it the medial triangle of a large one To prove that the altitude 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 with the vertex goes to the opposite side it 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 this will always, to prove that this will always be concurrent And you might say "wait! we How, how do how do we know that they are concurrent " Well, all you have to do is think about how they interact with the larger triangle How do these altitudes, what are these altitudes of 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, its core, 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, 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 Cause 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 is the way we constructed the larger triangle if they're going to be parallel So once again this is a perpendicular bisector So this whole reason, if you just give me any, any triangle I can take its altitudes and I know that its altitudes 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 altitude will be the perpendicular bisector for the larger triangle And we already know that the perpendicular bisector, is for any triangle, are concurrent They do intersect in exactly, in exactly one point