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Current time:0:00Total duration:10:01

Proof: Triangle altitudes are concurrent (orthocenter)

Video transcript

what I want to do in this video is to show that if we start with any arbitrary triangle this will be the arbitrary triangle that we're starting with that we can always make this the medial triangle of a larger triangle we say the medial triangle we mean that each of the vertices of these truck of this triangle will be the midpoint of the sides of a larger triangle I want 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 the side as a transversal of these two parallel lines or this line in 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's 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 you could always construct a line that's parallel to another line that 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 this 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 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 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 this with 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 thing 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 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 and 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 aside and an angle angle side angle congruence so these two are going to be congruent to each other then over here the on this inner triangle our original trying we'll decide this 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 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 sorry is 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 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 B e 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 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 E C so we've done what we wanted to do we've shown that if you start with any arbitrary triangle trying ADF triangle ADF we can construct a triangle BCE we can construct construct a triangle B C II so that ADF a DF is triangle BCE s 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 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 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 setup of this video is to show to prove that these will always to prove that these will always be concurrent and you might say wait well 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 to the larger triangle well this yellow altitude to the larger triangle remember these two yellow lines line ad and line C er parallel so if this is a 90 degree angle so it's it's core it's alternate interior angle is also going to be 90 degrees so this right over here is perpendicular to C E and it bisects C E because we know that ad 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 b.c.e 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 degree 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 should give me any any triangle I can take its altitudes and I know that it's 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 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 in exactly one point