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## High school geometry

### Course: High school geometry > Unit 4

Lesson 6: Proving relationships using similarity- Pythagorean theorem proof using similarity
- Exploring medial triangles
- Proof: Parallel lines divide triangle sides proportionally
- Prove theorems using similarity
- Proving slope is constant using similarity
- Proof: parallel lines have the same slope
- Proof: perpendicular lines have opposite reciprocal slopes

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# Exploring medial triangles

CCSS.Math:

The three midsegments (segments joining the midpoints of the sides) of a triangle form a medial triangle. Find out the properties of the midsegments, the medial triangle and the other 3 triangles formed in this way. Created by Sal Khan.

## Want to join the conversation?

- I did this problem using a theorem known as the midpoint theorem,which states that "the line segment joining the midpoint of any 2 sides of a triangle is parallel to the 3rd side and equal to half of it."(28 votes)
- There is a separate theorem called mid-point theorem. But it is actually nothing but similarity.(1 vote)

- Couldn't you just keep drawing out triangles over and over again like the Koch snowflake? Wouldn't it be fractal?(12 votes)
- Yes. A type of triangle like that is the Sierpinski Triangle.(9 votes)

- What is SAS similarity and what does it stand for? He mentioned it at3:00?(10 votes)
- Actually in similarity the ∆s are not congruent to each other but their sides are in proportion to

each other and angles correspond to each other .

Suppose we have ∆ABC and ∆PQR.

AB/PQ = BC/QR = AC/PR and angle A =angle P,angle B = angle Q and angle C = angle R.

Like congruency there are also test to prove that the ∆s are similar. For example SAS ,SSS, AA.

In SAS Similarity the two sides are in equal ratio and one angle is equal to another.(7 votes)

- it looks like the triangle is an equilateral triangle, so it makes 4 smaller equilateral triangles, but can you do the same to isoclines triangles?(4 votes)
- Yes, you could do that. Your starting triangle does not need to be equilateral or even isosceles, but you should be able to find the medial triangle for pretty much any triangle ABC. You can either believe me or you can look at the video again. While the original triangle in the video might look a bit like an equilateral triangle, it really is just a representative drawing. In the beginning of the video nothing is known or assumed about ABC, other than that it is a triangle, and consequently the conclusions drawn later on simply depend on ABC being a polygon with three vertices and three sides (i.e. some kind of triangle).(7 votes)

- what does that Medial Triangle look like to you?(5 votes)
- actually alec, its the tri force from zelda, which it more closely resembles than the harry potter thing(2 votes)

- Can Sal please make a video for the Triangle Midsegment Theorem? I'm really stuck on it and there's no video on here that quite matches up what I'm struggling with.(5 votes)
- why do his arrows look like smiley faces? lol(4 votes)
- Yes they do, don't they? lol(2 votes)

- Do medial triangles count as fractals because you can always continue the pattern?(4 votes)
- Medial triangles are considered as fractials because there is always most certianly going to be a pattern(2 votes)

- Does this work with any triangle, or only certain ones?(5 votes)
- The definition of "arbitrary" is "random".(0 votes)

- 2:50Sal says SAS similarity, but isn't it supposed to be SAS "congruency"?(1 vote)
- They are different things.

If two corresponding sides are congruent in different triangles and the angle measure between is the same, then the triangles are congruent. (SAS congruency)

If the ratio between one side and its corresponding counterpart is the same as another side and its corresponding counterpart, and the angles between them are the same, then the triangles are similar. (SAS similarity)

If the aforementioned ratio is equal to 1, then the triangles are congruent, so technically, congruency is a special case of similarity.(4 votes)

## Video transcript

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. 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.