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### Course: High school geometry > Unit 4

Lesson 3: Solving similar triangles# Solving similar triangles

Sal solves two problems where a missing side length is found by proving that triangles are similar and using this to find the measure. Created by Sal Khan.

## Want to join the conversation?

- Why do we need to do this? Will we be using this in our daily lives EVER?(75 votes)
**Similarity**and**proportional scaling**is quite useful in architecture, civil engineering, and many other professions. You will need similarity if you grow up to build or design cool things.(92 votes)

- Can someone sum this concept up in a nutshell? I'm having trouble understanding this.(41 votes)
- There are 5 ways to prove congruent triangles. SSS, SAS, AAS, ASA, and HL for right triangles. To prove similar triangles, you can use SAS, SSS, and AA.(16 votes)

- What is cross multiplying? (at3:38)(23 votes)
- Cross-multiplying is often used to solve proportions. As an example:

14/20 = x/100

Then multiply the numerator of the first fraction by the denominator of the second fraction:

1400 =

Then, multiply the denominator of the first fraction by the numerator of the second, and you will get:

1400 = 20x.

Solve by dividing both sides by 20. The answer is 70.(33 votes)

- I hate this. It makes me wanna commit crimes.(18 votes)
- what are alternate interiornangels(8 votes)
- Between two parallel lines, they are the angles on opposite sides of a transversal. Hope this helps!(6 votes)

- Bruh my brain can't process this.(9 votes)
- I still don't understand the concept. Can someone please help me?🥺(10 votes)
- I hate geometry. I dont get it.(9 votes)
- I learned this in my Geometry class. But why is it 5/8 instead of 5/3?(5 votes)
- Sal needed CA, and CA=CB+BA.(5 votes)

- in the 2nd question of this video, using c&d(componendo÷ndo), can't we figure out DE directly?(3 votes)
- We could, but it would be a little confusing and complicated. For instance, instead of using CD/CE at6:16, we could have made it something else that would give us the direct answer to DE. We could have put in DE + 4 instead of CE and continued solving. But it's safer to go the normal way.(7 votes)

## Video transcript

In this first problem
over here, we're asked to find out the length
of this segment, segment CE. And we have these
two parallel lines. AB is parallel to DE. And then, we have these two
essentially transversals that form these two triangles. So let's see what
we can do here. So the first thing that
might jump out at you is that this angle and this
angle are vertical angles. So they are going
to be congruent. The other thing that
might jump out at you is that angle CDE is an
alternate interior angle with CBA. So we have this transversal
right over here. And these are alternate
interior angles, and they are going
to be congruent. Or you could say that, if you
continue this transversal, you would have a corresponding
angle with CDE right up here and that this one's
just vertical. Either way, this
angle and this angle are going to be congruent. So we've established that
we have two triangles and two of the corresponding
angles are the same. And that by itself is enough
to establish similarity. We actually could show that
this angle and this angle are also congruent by
alternate interior angles, but we don't have to. So we already know
that they are similar. And actually, we
could just say it. Just by alternate
interior angles, these are also going
to be congruent. But we already know enough
to say that they are similar, even before doing that. So we already know
that triangle-- I'll color-code
it so that we have the same corresponding vertices. And that's really important--
to know what angles and what sides
correspond to what side so that you don't mess up
your, I guess, your ratios or so that you do know
what's corresponding to what. So we know triangle ABC
is similar to triangle-- so this vertex A corresponds
to vertex E over here. It's similar to vertex E. And
then, vertex B right over here corresponds to vertex D. EDC. Now, what does that do for us? Well, that tells us that the
ratio of corresponding sides are going to be the same. They're going to be
some constant value. So we have corresponding side. So the ratio, for example,
the corresponding side for BC is going to be DC. We can see it in just
the way that we've written down the similarity. If this is true, then BC is
the corresponding side to DC. So we know that the length
of BC over DC right over here is going to be equal to
the length of-- well, we want to figure
out what CE is. That's what we care about. And I'm using BC and DC
because we know those values. So BC over DC is going
to be equal to-- what's the corresponding side to CE? The corresponding
side over here is CA. It's going to be
equal to CA over CE. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC
is right over here. CA is 4. And now, we can
just solve for CE. Well, there's multiple ways
that you could think about this. You could cross-multiply, which
is really just multiplying both sides by both denominators. So you get 5 times
the length of CE. 5 times the length of CE is
equal to 3 times 4, which is just going to be equal to 12. And then we get CE is
equal to 12 over 5, which is the same thing
as 2 and 2/5, or 2.4. So this is going
to be 2 and 2/5. And we're done. We were able to use
similarity to figure out this side just
knowing that the ratio between the corresponding
sides are going to be the same. Now, let's do this
problem right over here. Let's do this one. Let me draw a little
line here to show that this is a
different problem now. This is a different problem. So in this problem, we need
to figure out what DE is. And we, once again, have these
two parallel lines like this. And so we know corresponding
angles are congruent. So we know that angle is going
to be congruent to that angle because you could view
this as a transversal. We also know that this
angle right over here is going to be congruent to
that angle right over there. Once again, corresponding
angles for transversal. And also, in both
triangles-- so I'm looking at triangle CBD
and triangle CAE-- they both share this angle up here. Once again, we could have
stopped at two angles, but we've actually shown that
all three angles of these two triangles, all three of
the corresponding angles, are congruent to each other. And once again, this is
an important thing to do, is to make sure that you
write it in the right order when you write your similarity. We now know that
triangle CBD is similar-- not congruent-- it is similar
to triangle CAE, which means that the ratio
of corresponding sides are going to be constant. So we know, for example,
that the ratio between CB to CA-- so let's
write this down. We know that the
ratio of CB over CA is going to be equal to
the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side is
going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we
can cross-multiply. We have 5CE. 5 times CE is
equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to
think about that, 6 and 2/5. Now, we're not done because
they didn't ask for what CE is. They're asking for just
this part right over here. They're asking for DE. So we know that this entire
length-- CE right over here-- this is 6 and 2/5. And so DE right over
here-- what we actually have to figure out--
it's going to be this entire length,
6 and 2/5, minus 4, minus CD right over here. So it's going to be 2 and 2/5. 6 and 2/5 minus 4
and 2/5 is 2 and 2/5. So we're done. DE is 2 and 2/5.