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### Course: Geometry (all content) > Unit 17

Lesson 1: Worked examples- Challenge problems: perimeter & area
- Challenging perimeter problem
- CA Geometry: Deductive reasoning
- CA Geometry: Proof by contradiction
- CA Geometry: More proofs
- CA Geometry: Similar triangles 1
- CA Geometry: More on congruent and similar triangles
- CA Geometry: Triangles and parallelograms
- CA Geometry: Area, pythagorean theorem
- CA Geometry: Area, circumference, volume
- CA Geometry: Pythagorean theorem, area
- CA Geometry: Exterior angles
- CA Geometry: Pythagorean theorem, compass constructions
- CA Geometry: Compass construction
- CA Geometry: Basic trigonometry
- CA Geometry: More trig
- CA Geometry: Circle area chords tangent
- Speed translation

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# CA Geometry: Similar triangles 1

10-14, similar triangles. Created by Sal Khan.

## Want to join the conversation?

- at about "7:30" what is the definition of similar triangles(17 votes)
- it means that the triangles are the same thing, but they might be bigger or smaller.(1 vote)

- weren't all the sides on the paralellogram on10:34the same?(4 votes)
- There are four paralellograms: Rhombus, Rhomboid, Rectangle, and a Square. The square is the only paralellogram must have four equal sides. While a Rhombus can have 4 equal lengths, for this problem, it is not a given that this is the case. We are looking for proof only that triangles inside the paralellogram are congruent. The answer to your question is the angles could be the same, either way, you should be able to get the right answer even if they were not.(2 votes)

- I don't get problem 13 at8:31

Can anyone help me?(1 vote)- For the 2 triangles to be similar the corresponding angles all have to have congruent measurements. For obvious reasons, angle DBE is the same on both tri. For the other 2 corresponding angles to be proven to be congruent, the 2 sides AC and DE must be parallel to eachother. This is true because the transversal of line AB would make the 2 corresponding angles of the triangle congruent ecause of the corresponding angles of the transversal are these angles. This is true for the other transversal as well.(4 votes)

- In problem #10, minute1:56, wouldn't the triangles be called RTP or PTR or just T for the one in purple, and the other to be called APT or TPA or just P. I thought that the major/main angle of the triangle needed to be in-between the other two letters when naming it?(3 votes)
- I'm not aware of defining a triangle based solely on the reference to the major angle.(0 votes)

- Do the arrows on line segments AC and BE at9:10represent parallel line segments?(2 votes)
- Yes, the arrows on line segments AC and BE do denote parallel line segments.(2 votes)

- Sal could have also used the SSS, couldn't he?(2 votes)
- You could do that, but first you would have to prove that the third side of the left triangle (RP) is equal to the third side of the right triangle (TA), and we aren't given that.

However, there is a rule when it comes to isosceles trapezoids which states that the diagonals on an isosceles trapezoids are always congruent. So if you knew that, then you would be able to state that, by this rule, RP must equal TA, and therefore you could use SSS.

However, if you already knew that RP equaled TA, then you wouldn't even need to prove that the triangles are congruent. Because "RP = TA" is the solution to the problem. So if you knew that, you would already be done.

I hope this helps.(2 votes)

- how do i use proofs like how do i write them and solve them(2 votes)
- is there proof? on how to do these thing thank >_<(2 votes)
- At5:09, he explains how the two scalene triangles with congruent bases cannot be similar. However, because they have congruent bases, wouldn't at least one of their angles be congruent to the other, the angle opposite the base? Thus, wouldn't the scalene triangles still be a viable option?(1 vote)
- On problem 11, the diagonals on choice B are not in any way perpendicular, nor a rhombus. Am i right? If not correct me, please.(1 vote)
- I forgot to mention that its also not a square. It's sort of a rectangle I guess.(1 vote)

## Video transcript

Actually, just right after
stopping that video, I realized a very simple way
of showing you that RP is congruent to TA, a little bit
more of a rigorous definition. If we can show that this
triangle right there, that one I drew in purple, and this
triangle right here are congruent, then we could make
a fairly reasonable argument that RP is going to be congruent
to TA, because they're essentially the
corresponding sides of the two congruent triangles. This congruent triangles
would be kind of flipping each other. So how can we make
that argument? Well, on the purple triangle,
this angle is going to be equal to this angle on
the yellow triangle. Actually, we got that from the
fact that this is an isosceles trapezoid, so the base angles
are going to be the same. They told us this is an
isosceles trapezoid, so we know that this side, right
there, is going to be congruent to this side. And then finally, they both
share this side right here. So we could use the argument--
once again, side, angle, side-- that the side, angle, and
side are congruent to this side, angle, and side. So you could say by SAS,
triangle TRP is congruent to triangle TAP. And if they're congruent, then
all of the corresponding sides are equal, so then TA
is congruent to RP. Once again, you didn't
have to do all that. It's a multiple choice test. But
I wanted to give you that. I felt bad I wasn't giving you
a more rigorous definition. A more rigorous proof. So anyway, problem number 11. A conditional statement
is shown below. If a quadrilateral has
perpendicular diagonals, then it is a rhombus. Fair enough. Which of the following is
a counterexample to the statement above? So they're saying, if it's
perpendicular diagonals, then it's a rhombus. So if we could find something
that has perpendicular diagonals that is not a rhombus,
then we have a counterexample. Then this would not be true. So let's find something with
perpendicular diagonals that is not a rhombus. Well, this one has perpendicular
diagonals. The diagonals are perpendicular
to each other, all 90-degree angles. And this is clearly
not a rhombus. This is like a kite. This is not parallel to this and
that is not parallel, so this is not a rhombus. So this is definitely
a counterexample. This one does have perpendicular
diagonals, but it's also a rhombus. So it's not a counterexample. It's just an example of what
they're trying to say. This has perpendicular
diagonals, it's a square, but a square is a subset
of rhombuses. So this is another example. And, of course, this
one does not have perpendicular diagonals. This is not a right angle. So A is the counterexample. Next question. Problem 12. Which triangles must
be similar? Two obtuse triangles. Well, obtuse just means that
they have two angles, one obtuse angle might look like
that where that is greater than 90 degrees there,
and the other obtuse might be super obtuse. It might be like that. And clearly these
aren't similar. Well, this angle is obviously
larger than that one. OK, so this is not similar. Similar means all the
angles are the same. So it's like congruent, but you
can scale them in size. That's how I think of them. Like this triangle. I'm trying to draw it so it
looks exactly the same. That triangle. Well no, but you can imagine
if I cut and pasted that, right? It would only be similar to
this triangle if I drew everything to scale. Because the sides are different
sizes, but all the angles are the same. That's what similar means. So let's see. Two scalene triangles with
congruent bases. Well no, that's not true. That's not similar. Let's say they share
the same base. One scalene triangle might
look like this. It might come out a little bit
and then go down like that. And the other scalene triangle
has the same base right there. The sides might be a little
closer to each other. So clearly, these two things
aren't similar. This angle is different
than that angle. All the angles are
different, so they're not similar triangles. So that's not right. Two right triangles. Do those have to be similar? Well, no. You could have a right triangle
that looks like this, where maybe the two sides
are equal, right? That's a 45-45-90 triangle. Or you could have something
like this where you have a 30-60-90 triangle. These clearly are not similar. All the angles are
not the same. They both have a 90-degree
angle. So I'm already guessing that D
is our right answer, but let's see how it works out. Two isosceles triangles with
congruent vertex angles. So I'm assuming when they say
congruent vertex angles, I'm assuming they mean all of the
angles are congruent. So two isosceles triangles. Let me think about
it a little bit. Oh, actually, I think what
they mean is angle in the middle when they say
vertex angle. If that's one of my isosceles
triangles. And isosceles triangle means
that that side is equal to that side and that angle
is equal to that angle. The vertex angle, I'm guessing,
they mean is this angle right there. So if I had another isosceles
triangle. Let's say maybe it's a
little bit smaller. It looks something like that. And their vertex angles
are the same. That angle is equal
to this angle. Well, if that angle is equal to
this angle and we know it's isosceles, so if we know it's
isosceles, that is equal to that, then that has to be equal
to that, we know that all the angles are the same. How do we know that this angle
is equal to this angle? Well, think about it. Whatever angle this us,
let's call this x. Let's call this angle
y and this angle y. We know that x plus 2y is equal
to 180, or that 2y is equal to 180 minus x, Or y is
equal to 90 minus x over 2. Now if this is x, And let's call
these z and z, So we know that x plus 2z is
equal to 180. All the angles in a triangle
have to add up to 180. Subtract x from both sides,
you get 2z is equal to 180 minus x. Divide by 2, you get z is equal
to 90 minus x over 2. So z and y are going to
be the same angles. So all the angles are the same,
so we're dealing with similar triangles. So choice D was definitely
correct. 13. OK. Which of the following facts
would be sufficient to prove that triangles ABC, that's the
big triangle, and triangle DBE, so that's a small
one, are similar? So we have to prove that all of
their angles are similar. I cannot even look at the
choices and I can guess where this is going. So we want to prove that
those are similar. So first of all, they share
the same angle. Angle ABC, this angle, is
the same as angle DBE. So they share that same angle. So we got one angle down. Now let's think about it. If we knew that this angle is
equal to that angle and that angle is equal to that
angle, we'd be done. And the best way to come to
that conclusion is if they told us that this and
this are parallel. I'm guessing that's where
they're going. Now I might have gone on a
completely wrong tangent. Because it those two are
parallel, then these two lines are transversals of
the parallel. So that angle and that angle
would be corresponding angles, so they would be congruent, and
then that angle and that angle would be corresponding
angles so they'd also be congruent. So if they told us that these
are parallel, we're done. These are definitely
similar triangles. And sure enough, choice C, they
tell us that AC and DE are parallel. These are parallel, that's
a transversal, this is a corresponding angle of that,
so they're congruent. This is a corresponding angle to
this, congruent, so all of the angles are congruent. So we have a similar triangle. Problem 14. OK. Parallelogram ABCD
is shown below. Fair enough. Parallelogram: that
tells us that the opposites sides are parallel. That's parallel to that, and
then this is parallel to that. And all of the choices got
clipped at the bottom, but I'll copy them over. Maybe I'll copy them
above the question. Well, let me see
what I can do. I think that's good enough. A little unconventional. OK. Parallelogram is shown below. They say which pair of triangles
can be established to be congruent to prove
that angle DAB is congruent to angle BCD? So DAB is this. Let me do it in another color. DAB is that angle, is
congruent to BCD. They want us to show
that those have the same angle measure. OK, and what do we
have to show? They say what pair of triangles
can be established to be congruent to prove that. OK, if these are both part of
two different congruent triangles and they are the
corresponding angles, then we know that they're congruent
and we'd be done. So let's see what they say. Triangle ADC and BCD. BCD has this angle in it. BCD does help us because it
has this angle in it, but triangle ADC does not have
this angle in it, right? Triangle ADC has this the
smaller angle in it. ADC doesn't involve this whole
thing, so that's not going to help us. Triangle AED, once again, does
not involve this larger angle, does not involve
the angle DAB. It only involves the little
smaller angle, so that's not going to help us. Triangle DAB. That looks good. That has this whole
angle in it. DAB. And then BCD. If we showed that that triangle
is congruent to this triangle right here,
I think we're done. That would be enough to show
that this angle is congruent to that angle, because they
would be the corresponding angles of a congruent
triangle. So I think C is where
we're going to go. Let's just look at choice D. DEC. Once again, triangle DEC. Let me make this point clear. Triangle DEC does not involve
either of the angles we care. It clearly does not involve
this angle, and it only involves part of this angle,
only this part. It doesn't involve this whole
angle, so that's not going to help us either. So the answer is C. Anyway, see you in
the next video.