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## Geometry (all content)

### Course: Geometry (all content) > Unit 4

Lesson 7: Altitudes# 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.

## Want to join the conversation?

- What does concurrent mean, when Sal says the altitudes are concurrent?(19 votes)
- Concurrent lines: Multiple lines that all intersect at a single point on a plane(5 votes)

- If it is a scalene triangle and the altitude of one of the sides forms two congruent angles, what would you say the reason is in you proof? I can't find anything on here but this about altitudes and congruent angles.(9 votes)
- The reason is Alternate interior angles of parrallel lines postulate (AIP postulate)(12 votes)

- So far (in geometry, i'm watching all the geometry vids in order) Sal has talked almost exclusively about triangles. When does he get into squares or even more complex stuff like nonagons?(5 votes)
- Triangles are the base shape in geometry. There are lots of theorems built around triangles. Triangles are the shape with the least sides. Also, every other polygon can be divided into triangles, because it is the base of all polygons. Triangle are very important to learn, especially in geometry, because they will be used in other areas of math too (so are circles too). Nonagons aren't that complex compared to triangles. Triangles are very important to know for a base of real, hard geometry. Nonagons, decagons, and other shapes with bizarre names just have more sides and angles. They really start to get boring after a while. Triangles, on the other hand, are so much more complex and we have lots to learn about them still.

Hope this helped!(14 votes)

- If you continue the altitudes (?) of the middle triangle, it looks like they intersect the larger triangle at some weird point, or is this just the way it is drawn? This seems like a very, very complicated way to find the middle of a triangle.(6 votes)
- The altitudes of the medial triangle end up being the perpendicular bisectors of the larger triangle so they won't necessarily go through any of its vertices. Perpendicular bisectors go through the midpoint of a side and are perpendicular to it but don't have to connect with a vertex.

The video didn't mention it explicitly but it ends up that the circumcenter of a triangle will be the orthocenter of its medial triangle.(7 votes)

- if O is the orthocentre of the trangle ABC and angle BAC is 80 degree then measure of angle BOC is(4 votes)
- If you know BAC, you know ACO and ABO (Draw the triangle to see why). With that, you know what OBC + OCB is (Draw the triangle to see how). And then, Finding BOC is extremely easy.(5 votes)

- So this is the easy way to find the orthocenter? Will this work every time? And how else can you find the orthocenter in simplest terms?(3 votes)
- just make altitudes and the point on which they intersect is orthocenter(3 votes)

- How do I find the altitude of a right triangle?(3 votes)
- An altitude is a perpendicular dropped on a different side from a vertex.

Therefore, the sides making the right angle will be altitudes themselves!!

The third altitude will fall on the hypotenuse.

therefore the orthocentre will be the vertice whose angle is 90 degrees(3 votes)

- 25) From a point in the interior of an equilateral triangle, altitudes to the 3 sides

are drawn. These altitudes have lengths 2, 6, and 4. Find the side length of this

triangle. (Refer to the diagram at the right)

A. B. C. D. E. NOTA(3 votes)- there was a guy in geometry named Viviani(1 vote)

- What is the difference between altitudes and medians?(1 vote)
- "A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.Altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side or an extension of the opposite side."(2 votes)

- What's the meaning of "concurrent"?(2 votes)
- Concurrent means things happening together. In math we use concurrent to mean we have multiple lines intersecting at one point. It isn't all that unusual for two lines to intersect (especially in a plane) but if three or more lines intersect at one point, we give that point a special name - a point of concurrency!(1 vote)

## Video transcript

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.