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).
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- 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
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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