If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:9:23

CCSS Math: HSG.SRT.B.5

What I want to do
in this video is see if we can identify similar
triangles here and prove to ourselves that they
really are similar, using some of the postulates
that we've set up. So over here, I
have triangle BDC. It's inside of triangle AEC. They both share this
angle right over there, so that gives us one angle. We need two to get
to angle-angle, which gives us similarity. And we know that these
two lines are parallel. We know if two
lines are parallel and we have a transversal
that corresponding angles are going to be congruent. So that angle is
going to correspond to that angle right over there. And we're done. We have one angle
in triangle AEC that is congruent to
another angle in BDC, and then we have
this angle that's obviously congruent to itself
that's in both triangles. So both triangles have a
pair of corresponding angles that are congruent, so
they must be similar. So we can write,
triangle ACE is going to be similar to
triangle-- and we want to get the letters
in the right order. So where the blue angle is
here, the blue angle there is vertex B. Then we go
to the wide angle, C, and then we go to the unlabeled
angle right over there, BCD. So we did that first one. Now let's do this
one right over here. This is kind of
similar, but it looks, just superficially looking at
it, that YZ is definitely not parallel to ST. So we
won't be able to do this corresponding
angle argument, especially because they didn't
even label it as parallel. And so you don't want to look
at things just by the way they look. You definitely want to
say, what am I given, and what am I not given? If these weren't
labeled parallel, we wouldn't be able
to make the statement, even if they looked parallel. One thing we do have is that
we have this angle right here that's common to
the inner triangle and to the outer
triangle, and they've given us a bunch of sides. So maybe we can use
SAS for similarity, meaning if we can show
the ratio of the sides on either side of
this angle, if they have the same ratio from the
smaller triangle to the larger triangle, then we
can show similarity. So let's go, and we have to go
on either side of this angle right over here. Let's look at the shorter side
on either side of this angle. So the shorter side
is two, and let's look at the shorter
side on either side of the angle for
the larger triangle. Well, then the shorter side
is on the right-hand side, and that's going to be XT. So what we want to
compare is the ratio between-- let me
write it this way. We want to see, is XY
over XT equal to the ratio of the longer side? Or if we're looking relative
to this angle, the longer of the two, not necessarily
the longest of the triangle, although it looks
like that as well. Is that equal to the ratio of
XZ over the longer of the two sides-- when you're
looking at this angle right here, on either
side of that angle, for the larger
triangle-- over XS? And it's a little confusing,
because we've kind of flipped which side, but I'm just
thinking about the shorter side on either side of
this angle in between, and then the longer side on
either side of this angle. So these are the shorter
sides for the smaller triangle and the larger triangle. These are the longer sides
for the smaller triangle and the larger triangle. And we see XY. This is two. XT is 3 plus 1 is 4. XZ is 3, and XS is 6. So you have 2 over
4, which is 1/2, which is the same thing as 3/6. So the ratio between
the shorter sides on either side of the
angle and the longer sides on either side of the
angle, for both triangles, the ratio is the same. So by SAS we know that the
two triangles are congruent. But we have to be careful on
how we state the triangles. We want to make sure we get
the corresponding sides. And I'm running
out of space here. Let me write it
right above here. We can write that triangle
XYZ is similar to triangle-- so we started up at X, which
is the vertex at the angle, and we went to the
shorter side first. So now we want to
start at X and go to the shorter side
on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at
this right over here. So in our larger triangle,
we have a right angle here, but we really know
nothing about what's going on with any of
these smaller triangles in terms of their actual angles. Even though this looks
like a right angle, we cannot assume it. And if we look at this smaller
triangle right over here, it shares one side with
the larger triangle, but that's not enough
to do anything. And then this triangle over
here also shares another side, but that also
doesn't do anything. So we really can't
make any statement here about any kind of similarity. So there's no similarity
going on here. There are some shared angles. This guy-- they both share
that angle, the larger triangle and the smaller triangle. So there could be a
statement of similarity we could make if we knew
that this definitely was a right angle. Then we could make some
interesting statements about similarity,
but right now, we can't really do anything as is. Let's try this one out,
this pair right over here. So these are the first
ones that we have actually separated out the triangles. So they've given us the three
sides of both triangles. So let's just figure
out if the ratios between corresponding
sides are a constant. So let's start with
the short side. So the short side here is 3. The shortest side here
is 9 square roots of 3. So we want to see whether the
ratio of 3 to 9 square roots of 3 is equal to the next
longest side over here, is 3 square roots of 3 over the
next longest side over here, which is 27. And then see if
that's going to be equal to the ratio
of the longest side. So the longest side
here is 6, and then the longest side over here
is 18 square roots of 3. So this is going to give
us-- let's see, this is 3. Let me do this in
a neutral color. So this becomes 1 over
3 square roots of 3. This becomes 1
over root 3 over 9, which seems like a
different number, but we want to be careful here. And then this right over here--
if you divide the numerator and denominator
by 6, this becomes a 1 and this becomes
3 square roots of 3. So 1 over 3 root 3 needs to
be equal to square root of 3 over 9, which needs to be equal
to 1 over 3 square roots of 3. At first they don't look
equal, but we can actually rationalize this
denominator right over here. We can show that 1 over
3 square roots of 3, if you multiply it
by square root of 3 over square root
of 3, this actually gives you in the
numerator square root of 3 over square root of 3 times
square root of 3 is 3, times 3 is 9. So these actually
are all the same. This is actually
saying, this is 1 over 3 root 3, which is the same thing
as square root of 3 over 9, which is this right
over here, which is the same thing
as 1 over 3 root 3. So actually, these
are similar triangles. So we can actually
say it, and I'll make sure I get the order right. So let's start with E,
which is between the blue and the magenta side. So that's between the
blue and the magenta side. That is H, right over here. I'll do it like this. Triangle E, and then I'll
go along the blue side, F. Actually, let me just
write it this way. Triangle EFG, we know
is similar to triangle-- So E is between the blue
and the magenta side. Blue and magenta side--
that is H. And then we go along the blue side to F,
go along the blue side to I, and then you went along
the orange side to G, and then you go along
the orange side to J. So triangle EFJ-- EFG is
similar to triangle HIJ by side-side-side similarity. They're not congruent sides. They all have just the same
ratio or the same scaling factor. Now let's do this last
one, right over here. Let's see. We have an angle that's
congruent to another angle right over there, and
we have two sides. And so it might be tempting
to use side-angle-side, because we have
side-angle-side here. And even the ratios look kind
of tempting, because 4 times 2 is 8. 5 times 2 is 10. But it's tricky
here, because they aren't the same
corresponding sides. In order to use side-angle-side,
the two sides that have the same
corresponding ratios, they have to be on
either side of the angle. So in this case, they are
on either side of the angle. In this case, the 4 is
on one side of the angle, but the 5 is not. So because if this
5 was over here, then we could make an
argument for similarity, but with this 5 not being on the
other side of the angle-- it's not sandwiching the
angle with the 4-- we can't use side-angle-side. And frankly, there's nothing
that we can do over here. So we can't make
some strong statement about similarity
for this last one.