CA Geometry: Similar Triangles 2 15-16, similar triangles
CA Geometry: Similar Triangles 2
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- We're on problem 15.
- It asks us, if triangle ABC and triangle XYZ are two
- triangles such that-- OK, let me draw these two triangle.
- Triangle ABC, maybe it looks something like that, and then
- we have triangle XYZ, and we want to prove
- that they are similar.
- Similar means that they look the same.
- They have a similar shape, but could be of different sizes.
- So essentially all of their angles are the same or all the
- ratios of their sides of the same.
- So XYZ might just be a smaller version.
- Maybe I should have drawn a bigger one.
- It looks like they want us to prove the
- triangles are similar.
- They tell us that the ratio of AB to XY-- let me color that.
- They say the ratio of AB to XY is equal to the
- ratio of BC to YZ.
- Which of the following would be sufficient to prove the
- triangles are similar?
- So there's a couple of times when you know that a triangle
- is similar, is if the ratio of all the sides are equal.
- Think about it, just giving two sides isn't enough,
- because even though I drew them so that they're similar,
- it could look something like this.
- Maybe the ratio here is 2:1.
- So maybe AB and XY, maybe that's a 2:1.
- Maybe that's 2 and that's 1 like that.
- The ratio is 2:1.
- And maybe BC to YZ is also 2:1.
- But there's nothing that tells us that that can't
- open up like this.
- It could be like a 2 and a 1 ratio.
- Or actually even better, I don't want to draw them
- similar anymore.
- I want to draw them some unsimilar.
- Maybe BC comes in like that maybe YZ goes out like that.
- And so just knowing that the ratio of two of the sides are
- the same, that alone doesn't tell you that you're dealing
- with a similar triangle.
- These two triangles would be very different.
- These are definitely not similar triangles.
- In fact, all of the angles would be different.
- Now, how do you prove that something is similar?
- Well, actually I attempted this problem about two minutes
- ago and it stumped me, because I think, if I'm right, that
- there's a mistake in the problem.
- In geometry class, they always teach you that if you know
- that the ratio of two of the sides are equal and that the
- angle between them are equal, then that's sufficient to say
- that it's a similar triangle.
- Or that you can make the conclusion of the ratio of AC
- to XZ would also be equal.
- So that's the answer they're probably looking for.
- That if you have the ratio of AB to XY, you have the ratio
- of BC to YZ, then you want the angle in between them to say,
- OK, these are similar triangles.
- So you'd have to know that angle B is
- congruent to angle Y.
- And that's choice B.
- I don't argue with the fact that angle B being congruent
- to angle Y is sufficient to prove that these
- triangles are similar.
- I just think that some of these other things are
- actually sufficient to prove the triangles are similar.
- They don't teach you the tools in geometry, but you can even
- think about it intuitively.
- Once you learn the law of cosines in trigonometry, which
- you're not that far away from right now, you'll see that any
- of the other angles, if you know the corresponding angles
- are congruent, you would be all set to know that these are
- going to be similar triangles.
- For example, if you know that angle C is congruent to angle
- Z, and you could even think about it.
- Let's say if you knew that angle C was congruent to angle
- Z, there's no other way to draw this triangle versus just
- having a similar triangle.
- So, for example, in this one right here, this would be
- angle C and this would be angle Z, and clearly, these
- two aren't congruent.
- If you think about it, when you say that this angle is
- congruent to this angle, you're constraining the shape
- of this line right here.
- You're constraining the direction that it goes in.
- If you forced this angle to be the same as this, let's say
- that you said this angle is equal to that angle, then you
- would have to go out in this direction out there, and then
- you would break what you originally said, that the
- ratio of that to that is equal.
- So just think about it in terms of when you add another
- angle, it actually does constrain the triangle to
- being similar.
- And the same thing is true, angle A and angle X.
- If you know that angle A is congruent to angle X, that
- also constrains the triangle to being similar.
- The only one that I know definitely isn't sufficient is
- angle X being congruent to angle Y.
- That's useless.
- Let's see, angle X congruent to angle Y.
- Yeah, that doesn't help you.
- That tells you that this one is an isosceles triangle
- because these two base angles would be the same, so that
- would be the same as that, but it doesn't tell you anything
- about how does this relate to that.
- So even though the answer they're looking for is
- probably B, I would say that A and C are also sufficient.
- Next question.
- Question 16.
- And send me a note if you think I'm missing something,
- because I thought about it a little bit.
- All right, 16.
- In parallelogram FGHI-- let me draw parallelogram FGHI.
- We have one side.
- We come down at a 45-degree angle.
- We have the other side.
- Let's say we go up at a 45-degree angle.
- I can do it from here.
- Close enough.
- Got my parallelogram.
- Parallelogram FGHI.
- So that F, that's a G, that's an H, and that's an I.
- Diagonals IG and FH are drawn and intersect at point M.
- So let me draw IG and FH.
- OK, these are lines.
- Fair enough.
- And they intersect at point M.
- Which of the following statements must be true?
- Triangle FGI must be an obtuse triangle.
- And that means that one of the angles in the triangle is more
- than 90 degrees.
- I mean, just the way I drew it here, which is a completely
- legitimate way to draw it, all of these lines are parallel.
- We see that that doesn't have to be the case, because just
- the way I drew it, all these angles are
- less than 90 degrees.
- So that's definitely not the case.
- Triangle HIG, must be an acute triangle.
- Let's think about it a little bit.
- Let's try to prove it by contradiction.
- Let's say that one of the sides was
- more than 90 degrees.
- So this is just the way I happened to
- draw it right here.
- But if you think about it, I could have drawn it like this.
- So the way I drew it, HIG is in an acute triangle.
- All of these sides are less than 90 degrees.
- But I could have drawn this the complete opposite way.
- Let me draw it that way actually.
- I'll do it quick and dirty.
- So what if I drew it like this?
- What if I drew our parallelogram like this?
- The direction that I drew it was arbitrary.
- So now this is F, G, H and I, and they're talking about
- triangle HIG.
- Well, clearly, in this case, angle H right here, this is an
- obtuse angle.
- That is greater than 90 degrees.
- So HIG would be an obtuse triangle.
- So there's nothing that prevents us from drawing the
- parallelogram like this, which would make
- this an obtuse triangle.
- So that triangle does not have to be an acute triangle, so B
- doesn't work.
- All right, what else do they say?
- Triangle FMG must be congruent to triangle HMG.
- Well that's congruent.
- So that would mean that all the sides are equal and
- everything, and that's clearly not true.
- They share this side right here, but this side up here
- could be a lot longer than that side, which would make
- this side here a lot longer than that side, so that's
- definitely not true.
- So, D is probably the answer, but let's see if
- we can prove it.
- Triangle GMH must be congruent to IMF.
- So let me do that in two different colors.
- So GMH, that's that one right there, must be congruent to
- IMF, that's that one right there.
- Let's see.
- First of all, we know that this angle and that angle are
- going to be congruent.
- They are opposite angles or, as they say, vertical angles.
- The reason I don't like using vertical angles is because
- this angle and this angle are also opposite angles, and to
- call them vertical angles is bizarre because they're not
- vertical with each other.
- Anyway, that's just my problem with the notation.
- OK, so where were we?
- We wanted to prove that that triangle and that
- triangle are congruent.
- This angle is definitely equal to that angle, and know that
- this line is parallel to this line, right?
- So we can view each of these crossing lines as transversals
- will between two parallel lines.
- That's a parallel line, that's a parallel line, so this is a
- transversal between two parallel lines.
- Then this angle is going to be congruent to that angle, and
- that's because I think the word is opposite internal, or
- alternate interior angles, or something like that.
- And it makes sense.
- You could imagine drawing the transversal either way and
- then the angles would change, but they are always going to
- be congruent.
- And by the same argument, this angle is going to be congruent
- to that angle.
- So we know that all of the angles are the same, and now
- we have to make some type of argument that all of the sides
- are the same.
- Really, if we could just prove that one of the sides are the
- same, then we know that all of the sides are the same.
- Well, we know that it's a parallelogram.
- So in order for all of the sides to be parallel, the
- opposite sides have to be the same length.
- I'll leave you to think about why that is.
- But if all the opposite sides are the same length, we know
- that this side is equivalent to this side.
- So, you know, we could use one of those geometry postulates
- that they-- angle, side, angle, right?
- We haven an angle, a side and an angle, and an angle, a side
- and an angle, and say, OK, that's enough for us to show
- that these are congruent triangles.
- So the answer is D.
- Anyway, see you in the next video.
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