So I've got an arbitrary triangle here.
We'll call it triangle ABC.
And what I want to do is look at the midpoints
of each of the sides of ABC.
So this is the midpoint of one of the sides, of side BC.
Let's call that point D. Let's call this midpoint E. And let's
call this midpoint right over here F.
And since it's the midpoint, we know
that the distance between BD is equal to the distance from D
to C. So this distance is equal to this distance.
We know that AE is equal to EC, so this distance
is equal to that distance.
And we know that AF is equal to FB,
so this distance is equal to this distance.
Instead of drawing medians going from these midpoints
to the vertices, what I want to do is
I want to connect these midpoints and see what happens.
So if I connect them, I clearly have three points.
So if you connect three non-linear points like this,
you will get another triangle.
And this triangle that's formed from the midpoints of the sides
of this larger triangle-- we call this a medial triangle.
And that's all nice and cute by itself.
But what we're going to see in this video
is that the medial triangle actually
has some very neat properties.
What we're actually going to show
is that it divides any triangle into four smaller triangles
that are congruent to each other,
that all four of these triangles are identical to each other.
And they're all similar to the larger triangle.
And you could think of them each as having
1/4 of the area of the larger triangle.
So let's go about proving it.
So first, let's focus on this triangle down
here, triangle CDE.
And it looks similar to the larger triangle,
to triangle CBA.
But let's prove it to ourselves.
So one thing we can say is, well, look, both of them
share this angle right over here.
Both the larger triangle, triangle CBA, has this angle.
And the smaller triangle, CDE, has this angle.
So they definitely share that angle.
And then let's think about the ratios of the sides.
We know that the ratio of CD to CB is equal to 1 over 2.
This is 1/2 of this entire side, is equal to 1 over 2.
And that's the same thing as the ratio of CE to CA.
CE is exactly 1/2 of CA, because E is the midpoint.
It's equal to CE over CA.
So we have an angle, corresponding angles that
are congruent, and then the ratios
of two corresponding sides on either side of that angle
are the same.
CD over CB is 1/2, CE over CA is 1/2, and the angle in between
is congruent.
So by SAS similarity, we know that triangle CDE
is similar to triangle CBA.
And just from that, you can get some interesting results.
Because then we know that the ratio
of this side of the smaller triangle to the longer triangle
is also going to be 1/2.
Because the other two sides have a ratio of 1/2,
and we're dealing with similar triangles.
So this is going to be 1/2 of that.
And we know 1/2 of AB is just going to be the length of FA.
So we know that this length right over here
is going to be the same as FA or FB.
And we get that straight from similar triangles.
Because these are similar, we know that DE over BA
has got to be equal to these ratios,
the other corresponding sides, which is equal to 1/2.
And so that's how we got that right over there.
Now let's think about this triangle up here.
We could call it BDF.
So first of all, if we compare triangle
BDF to the larger triangle, they both share
this angle right over here, angle ABC.
They both have that angle in common.
And we're going to have the exact same argument.
You can just look at this diagram.
And you know that the ratio of BA-- let me do it this way.
The ratio of BF to BA is equal to 1/2,
which is also the ratio of BD to BC.
The ratio of this to that is the same
as the ratio of this to that, which is 1/2.
Because BD is 1/2 of this whole length.
BF is 1/2 of that whole length.
And so you have corresponding sides
have the same ratio on the two triangles,
and they share an angle in between.
So once again, by SAS similarity,
we know that triangle-- I'll write it
this way-- DBF is similar to triangle CBA.
And once again, we use this exact same kind of argument
that we did with this triangle.
Well, if it's similar, the ratio of all the corresponding sides
have to be the same.
And that ratio is 1/2.
So the ratio of this side to this side,
the ratio of FD to AC, has to be 1/2.
Or FD has to be 1/2 of AC.
And 1/2 of AC is just the length of AE.
So that is just going to be that length right over there.
I think you see where this is going.
And also, because it's similar, all of the corresponding angles
have to be the same.
And we know that the larger triangle
has a yellow angle right over there.
So we'd have that yellow angle right over here.
And this triangle right over here
was also similar to the larger triangle.
So it will have that same angle measure up here.
We already showed that in this first part.
So now let's go to this third triangle.
I think you see the pattern.
I'm sure you might be able to just pause this video
and prove it for yourself.
But we see that the ratio of AF over AB
is going to be the same as the ratio of AE
over AC, which is equal to 1/2.
So we have two corresponding sides where the ratio is 1/2,
from the smaller to larger triangle.
And they share a common angle.
They share this angle in between the two sides.
So by SAS similarity-- this is getting repetitive
now-- we know that triangle EFA is similar to triangle CBA.
And so the ratio of all of the corresponding sides
need to be 1/2.
So the ratio of FE to BC needs to be 1/2,
or FE needs to be 1/2 of that, which is just the length of BD.
So this is just going to be that length right over there.
And you can also say that since we've
shown that this triangle, this triangle, and this triangle--
we haven't talked about this middle one
yet-- they're all similar to the larger triangle.
So they're also all going to be similar to each other.
So they're all going to have the same corresponding angles.
So if the larger triangle had this yellow angle here,
then all of the triangles are going
to have this yellow angle right over there.
And if the larger triangle had this blue angle
right over here, then in the corresponding vertex,
all of the triangles are going to have that blue angle.
All of the ones that we've shown are similar.
We haven't thought about this middle triangle just yet.
And of course, if this is similar to the whole,
it'll also have this angle at this vertex
right over here, because this corresponds to that vertex,
based on the similarity.
So that's interesting.
Now let's compare the triangles to each other.
We've now shown that all of these triangles
have the exact same three sides.
Has this blue side-- or actually, this one-mark side,
this two-mark side, and this three-mark side.
One mark, two mark, three mark.
And that even applies to this middle triangle
right over here.
So by side-side-side congruency, we now know--
and we want to be careful to get our corresponding sides right--
we now know that triangle CDE is congruent to triangle DBF.
I want to get the corresponding sides.
I'm looking at the colors.
I went from yellow to magenta to blue, yellow, magenta,
to blue, which is going to be congruent to triangle
EFA, which is going to be congruent to this triangle
in here.
But we want to make sure that we're
getting the right corresponding sides here.
So to make sure we do that, we just
have to think about the angles.
So we know-- and this is interesting--
that because the interior angles of a triangle
add up to 180 degrees, we know this magenta
angle plus this blue angle plus this yellow angle equal 180.
Here, we have the blue angle and the magenta angle,
and clearly they will all add up to 180.
So you must have the blue angle.
The blue angle must be right over here.
Same argument-- yellow angle and blue angle,
we must have the magenta angle right over here.
They add up to 180.
So this must be the magenta angle.
And then finally, magenta and blue-- this
must be the yellow angle right over there.
And so when we wrote the congruency here,
we started at CDE.
We went yellow, magenta, blue.
So over here, we're going to go yellow, magenta, blue.
So it's going to be congruent to triangle FED.
And so that's pretty cool.
We just showed that all three, that this triangle,
this triangle, this triangle, and that triangle
are congruent.
And also, we can look at the corresponding--
and that they all have ratios relative to-- they're
all similar to the larger triangle, to triangle ABC.
And that the ratio between the sides is 1 to 2.
And also, because we've looked at corresponding angles,
we see, for example, that this angle
is the same as that angle.
So if you viewed DC or if you viewed BC
as a transversal, all of a sudden
it becomes pretty clear that FD is going to be parallel to AC,
because the corresponding angles are congruent.
So this is going to be parallel to that right over there.
And then you could use that same exact argument
to say, well, then this side, because once again,
corresponding angles here and here-- you
could say that this is going to be
parallel to that right over there.
And then finally, you make the same argument over here.
I want to make sure I get the right corresponding angles.
You have this line and this line.
And this angle corresponds to that angle.
They're the same.
So this DE must be parallel to BA.
So that's another neat property of this medial triangle,
[? I thought. ?] All of these things
just jump out when you just try to do something fairly simple
with a triangle.