High school geometry
- Pythagorean theorem proof using similarity
- Exploring medial triangles
- Proof: Parallel lines divide triangle sides proportionally
- Prove theorems using similarity
- Proving slope is constant using similarity
- Proof: parallel lines have the same slope
- Proof: perpendicular lines have opposite reciprocal slopes
Exploring medial triangles
The three midsegments (segments joining the midpoints of the sides) of a triangle form a medial triangle. Find out the properties of the midsegments, the medial triangle and the other 3 triangles formed in this way. Created by Sal Khan.
Want to join the conversation?
- I did this problem using a theorem known as the midpoint theorem,which states that "the line segment joining the midpoint of any 2 sides of a triangle is parallel to the 3rd side and equal to half of it."(28 votes)
- There is a separate theorem called mid-point theorem. But it is actually nothing but similarity.(1 vote)
- Couldn't you just keep drawing out triangles over and over again like the Koch snowflake? Wouldn't it be fractal?(12 votes)
- Yes. A type of triangle like that is the Sierpinski Triangle.(9 votes)
- What is SAS similarity and what does it stand for? He mentioned it at3:00?(10 votes)
- Actually in similarity the ∆s are not congruent to each other but their sides are in proportion to
each other and angles correspond to each other .
Suppose we have ∆ABC and ∆PQR.
AB/PQ = BC/QR = AC/PR and angle A =angle P,angle B = angle Q and angle C = angle R.
Like congruency there are also test to prove that the ∆s are similar. For example SAS ,SSS, AA.
In SAS Similarity the two sides are in equal ratio and one angle is equal to another.(7 votes)
- it looks like the triangle is an equilateral triangle, so it makes 4 smaller equilateral triangles, but can you do the same to isoclines triangles?(4 votes)
- Yes, you could do that. Your starting triangle does not need to be equilateral or even isosceles, but you should be able to find the medial triangle for pretty much any triangle ABC. You can either believe me or you can look at the video again. While the original triangle in the video might look a bit like an equilateral triangle, it really is just a representative drawing. In the beginning of the video nothing is known or assumed about ABC, other than that it is a triangle, and consequently the conclusions drawn later on simply depend on ABC being a polygon with three vertices and three sides (i.e. some kind of triangle).(7 votes)
- what does that Medial Triangle look like to you?(5 votes)
- actually alec, its the tri force from zelda, which it more closely resembles than the harry potter thing(2 votes)
- Can Sal please make a video for the Triangle Midsegment Theorem? I'm really stuck on it and there's no video on here that quite matches up what I'm struggling with.(5 votes)
- Do medial triangles count as fractals because you can always continue the pattern?(4 votes)
- Medial triangles are considered as fractials because there is always most certianly going to be a pattern(2 votes)
- Does this work with any triangle, or only certain ones?(5 votes)
- The definition of "arbitrary" is "random".(0 votes)
- why do his arrows look like smiley faces? lol(2 votes)
- Yes they do, don't they? lol(2 votes)
- 2:50Sal says SAS similarity, but isn't it supposed to be SAS "congruency"?(1 vote)
- They are different things.
If two corresponding sides are congruent in different triangles and the angle measure between is the same, then the triangles are congruent. (SAS congruency)
If the ratio between one side and its corresponding counterpart is the same as another side and its corresponding counterpart, and the angles between them are the same, then the triangles are similar. (SAS similarity)
If the aforementioned ratio is equal to 1, then the triangles are congruent, so technically, congruency is a special case of similarity.(4 votes)
So I've got an arbitrary triangle here. We'll call it triangle ABC. And what I want to do is look at the midpoints of each of the sides of ABC. So this is the midpoint of one of the sides, of side BC. Let's call that point D. Let's call this midpoint E. And let's call this midpoint right over here F. And since it's the midpoint, we know that the distance between BD is equal to the distance from D to C. So this distance is equal to this distance. We know that AE is equal to EC, so this distance is equal to that distance. And we know that AF is equal to FB, so this distance is equal to this distance. Instead of drawing medians going from these midpoints to the vertices, what I want to do is I want to connect these midpoints and see what happens. So if I connect them, I clearly have three points. So if you connect three non-linear points like this, you will get another triangle. And this triangle that's formed from the midpoints of the sides of this larger triangle-- we call this a medial triangle. And that's all nice and cute by itself. But what we're going to see in this video is that the medial triangle actually has some very neat properties. What we're actually going to show is that it divides any triangle into four smaller triangles that are congruent to each other, that all four of these triangles are identical to each other. And they're all similar to the larger triangle. And you could think of them each as having 1/4 of the area of the larger triangle. So let's go about proving it. So first, let's focus on this triangle down here, triangle CDE. And it looks similar to the larger triangle, to triangle CBA. But let's prove it to ourselves. So one thing we can say is, well, look, both of them share this angle right over here. Both the larger triangle, triangle CBA, has this angle. And the smaller triangle, CDE, has this angle. So they definitely share that angle. And then let's think about the ratios of the sides. We know that the ratio of CD to CB is equal to 1 over 2. This is 1/2 of this entire side, is equal to 1 over 2. And that's the same thing as the ratio of CE to CA. CE is exactly 1/2 of CA, because E is the midpoint. It's equal to CE over CA. So we have an angle, corresponding angles that are congruent, and then the ratios of two corresponding sides on either side of that angle are the same. CD over CB is 1/2, CE over CA is 1/2, and the angle in between is congruent. So by SAS similarity, we know that triangle CDE is similar to triangle CBA. And just from that, you can get some interesting results. Because then we know that the ratio of this side of the smaller triangle to the longer triangle is also going to be 1/2. Because the other two sides have a ratio of 1/2, and we're dealing with similar triangles. So this is going to be 1/2 of that. And we know 1/2 of AB is just going to be the length of FA. So we know that this length right over here is going to be the same as FA or FB. And we get that straight from similar triangles. Because these are similar, we know that DE over BA has got to be equal to these ratios, the other corresponding sides, which is equal to 1/2. And so that's how we got that right over there. Now let's think about this triangle up here. We could call it BDF. So first of all, if we compare triangle BDF to the larger triangle, they both share this angle right over here, angle ABC. They both have that angle in common. And we're going to have the exact same argument. You can just look at this diagram. And you know that the ratio of BA-- let me do it this way. The ratio of BF to BA is equal to 1/2, which is also the ratio of BD to BC. The ratio of this to that is the same as the ratio of this to that, which is 1/2. Because BD is 1/2 of this whole length. BF is 1/2 of that whole length. And so you have corresponding sides have the same ratio on the two triangles, and they share an angle in between. So once again, by SAS similarity, we know that triangle-- I'll write it this way-- DBF is similar to triangle CBA. And once again, we use this exact same kind of argument that we did with this triangle. Well, if it's similar, the ratio of all the corresponding sides have to be the same. And that ratio is 1/2. So the ratio of this side to this side, the ratio of FD to AC, has to be 1/2. Or FD has to be 1/2 of AC. And 1/2 of AC is just the length of AE. So that is just going to be that length right over there. I think you see where this is going. And also, because it's similar, all of the corresponding angles have to be the same. And we know that the larger triangle has a yellow angle right over there. So we'd have that yellow angle right over here. And this triangle right over here was also similar to the larger triangle. So it will have that same angle measure up here. We already showed that in this first part. So now let's go to this third triangle. I think you see the pattern. I'm sure you might be able to just pause this video and prove it for yourself. But we see that the ratio of AF over AB is going to be the same as the ratio of AE over AC, which is equal to 1/2. So we have two corresponding sides where the ratio is 1/2, from the smaller to larger triangle. And they share a common angle. They share this angle in between the two sides. So by SAS similarity-- this is getting repetitive now-- we know that triangle EFA is similar to triangle CBA. And so the ratio of all of the corresponding sides need to be 1/2. So the ratio of FE to BC needs to be 1/2, or FE needs to be 1/2 of that, which is just the length of BD. So this is just going to be that length right over there. And you can also say that since we've shown that this triangle, this triangle, and this triangle-- we haven't talked about this middle one yet-- they're all similar to the larger triangle. So they're also all going to be similar to each other. So they're all going to have the same corresponding angles. So if the larger triangle had this yellow angle here, then all of the triangles are going to have this yellow angle right over there. And if the larger triangle had this blue angle right over here, then in the corresponding vertex, all of the triangles are going to have that blue angle. All of the ones that we've shown are similar. We haven't thought about this middle triangle just yet. And of course, if this is similar to the whole, it'll also have this angle at this vertex right over here, because this corresponds to that vertex, based on the similarity. So that's interesting. Now let's compare the triangles to each other. We've now shown that all of these triangles have the exact same three sides. Has this blue side-- or actually, this one-mark side, this two-mark side, and this three-mark side. One mark, two mark, three mark. One mark, two mark, three mark. And that even applies to this middle triangle right over here. So by side-side-side congruency, we now know-- and we want to be careful to get our corresponding sides right-- we now know that triangle CDE is congruent to triangle DBF. I want to get the corresponding sides. I'm looking at the colors. I went from yellow to magenta to blue, yellow, magenta, to blue, which is going to be congruent to triangle EFA, which is going to be congruent to this triangle in here. But we want to make sure that we're getting the right corresponding sides here. So to make sure we do that, we just have to think about the angles. So we know-- and this is interesting-- that because the interior angles of a triangle add up to 180 degrees, we know this magenta angle plus this blue angle plus this yellow angle equal 180. Here, we have the blue angle and the magenta angle, and clearly they will all add up to 180. So you must have the blue angle. The blue angle must be right over here. Same argument-- yellow angle and blue angle, we must have the magenta angle right over here. They add up to 180. So this must be the magenta angle. And then finally, magenta and blue-- this must be the yellow angle right over there. And so when we wrote the congruency here, we started at CDE. We went yellow, magenta, blue. So over here, we're going to go yellow, magenta, blue. So it's going to be congruent to triangle FED. And so that's pretty cool. We just showed that all three, that this triangle, this triangle, this triangle, and that triangle are congruent. And also, we can look at the corresponding-- and that they all have ratios relative to-- they're all similar to the larger triangle, to triangle ABC. And that the ratio between the sides is 1 to 2. And also, because we've looked at corresponding angles, we see, for example, that this angle is the same as that angle. So if you viewed DC or if you viewed BC as a transversal, all of a sudden it becomes pretty clear that FD is going to be parallel to AC, because the corresponding angles are congruent. So this is going to be parallel to that right over there. And then you could use that same exact argument to say, well, then this side, because once again, corresponding angles here and here-- you could say that this is going to be parallel to that right over there. And then finally, you make the same argument over here. I want to make sure I get the right corresponding angles. You have this line and this line. And this angle corresponds to that angle. They're the same. So this DE must be parallel to BA. So that's another neat property of this medial triangle, [? I thought. ?] All of these things just jump out when you just try to do something fairly simple with a triangle.