Learn about CA Geometry: Similar Triangles

Khan Academy Presents: CA Geometry: 15-16, similar triangles

We’re on problem 15, it asks us, if triangle ABC and triangle XYZ or two triangles such that, okay let me draw these two triangles. So, triangle ABC, maybe it look something like that. ABC and then we have triangle XYZ and we want to prove that they are similar. So, similar means that they look the same, they have the similar shape but they could be of different sizes so essentially all of their angles are the same, all the ratios of their side are the same so XYZ might just be a smaller version maybe I should have drawn a bigger one, XYZ. It looks like they want us to prove the triangles are similar. They tell us that side, the ratio of AB to XY, let me color that. They said the ratio of AB to XY is equal to the ratio of BC to YZ. Which of the following would be sufficient to prove that triangles are similar? So, there’s a couple of times when you know that a triangle is similar. Is it the ratio of all the sides are equal? Think about just giving two sides isn’t enough because eventhough I drew them so that they are similar, it could look something like this. Maybe the ratio here is, I don’t know two to one. So, maybe AB and XY maybe that’s a two to one, maybe that’s two and that’s one like that, right. The ratio is 2 to 1 and maybe BC to YZ is also 2 to 1. For y, there’s nothing to tell us that it can't open up like this. It could be like a two and a one ratio right or actually even better I don’t want to draw them similar anymore, I want to draw them unsimilar. Maybe BC comes in like that, maybe YZ goes out like that and so just knowing that the ratio of two of the triangles are two of the sides are the same, that alone doesn’t tell you that you’re dealing with the 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 is 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 is a similar triangle or that you can make the conclusion that 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 okay these are similar triangles so you have to know that angle B is congruent to angle Y and that is choice B. Now, 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, then they don’t teach you the tools in geometry but you can even think about it intuitively. I mean once you learn the law of cosines and trigonometry which you are not that far away from right now, you’ll see that any of the other angles if you know that the corresponding angles are congruent, you would be all set to know that these are going to be similar triangles. So if you know that angle, for example if you know that angle C is congruent to angle C and you can even think about it. Let’s say, if you need an angle C was congruent to angle Z there is no other way to draw this triangle at versus just having a similar triangle. So example, see 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 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, right. You’re constraining the direction that goes in. If you force this angle to be the same as 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’ve 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 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. That just tells you that some type of an isosceles, you see angle X congruent to angle Y, yeah that doesn’t help you. I mean that tells you that this one is an isosceles triangle because you have this 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, eventhough the answer is they’re looking for it’s probably B, I would say that A and C were 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 a parallelogram FGHI and one side and let’s see we come down at a 45-degree angle. If the other side, now let’s say we go up in a 45-degree angle, close enough got my parallelogram. Parallelogram FGHI, so that’s 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. Okay, these are lines IG and FH, fair enough and they intersect at point M. Which of the following statements must be true? See triangle FGI must be an obtuse triangle and that means that one of the angles in the triangle is not more than 90 degrees. I mean just the way I drew it here which is completely legitimate way to draw and 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 of these angles are less than 90 degrees so that’s definitely not the case. Triangle HIG must be an acute triangle. So, 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 happen to draw it right here but if you think about it I could have drawn it like this. So, the way I drew an HIG is an acute triangle. All of these sides are less than 90 degrees but I could’ve 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, right? The direction that I drew it is arbitrary. So now, this is FGHI and they’re telling about triangle HIG right. Well clearly in this case, angle H right here, this is an obtuse angle. That is greater than 90 degrees so there is nothing and so the HIG would be an obtuse triangle. So, there is 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 enough. I mean 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. First of all, we know that this angle and that angle are going to be congruent to their opposite angles or they say vertical angles because I don’t like using vertical angles because this angle and this angle are also opposite angles but, and they call them vertical angles is—because it’s not vertical with each other, anyway that’s just my problem with the notation. We want to prove that that triangle is congruent. And this angle is definitely equal to that angle and we know that this line is parallel to this line right. So, we could view each of these crossing lines as transversal between two parallel lines. All right that’s parallel line. So, this is the transversal between two parallel lines then this angle is going to be congruent to that angle and that’s because well I think the word is opposite internal or alternate interior angles or something like that and it make sense. You could imagine drawing the transversal either way then the angles would change but they’re 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 you can just prove that one of the sides are the same and we know that all of the sides are the same. Well, we know that’s a parallelogram. So, in order for all of the sides to be parallel, the opposite sides have to be the same length. And I’ll leave you and think about why that is but if all the sides are the same or 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 you know angle side angle. Right, we have an angle side and an angle. Okay, that’s enough for us to show that this is congruent, these are congruent triangles so the answer is D. Anyway, see you in the next video.