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Current time:0:00Total duration:11:33

CCSS.Math:

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 medial triangle and that's all nice and cute by itself but 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 are identical to each other and they're all similar to the larger triangle and you could think of them each has having one-fourth 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 it looks similar to the larger triangle - triangle C ba 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 shares 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 we know that the ratio of C D to C be the ratio CD to CB to see B is equal to one over two this is half of this entire side is equal to one over two and that's the same thing as the ratio of C to C a c e is exactly half of CA it's because E is the midpoint it's equal to C E over C a so we have an angle and corresponding angles that are congruent and then the ratios of two corresponding sides on either side of that angle are the same see D over C B is 1/2 C E over C a is 1/2 and the angle in between is congruent so by SAS so by SAS similarity SAS similarity similarity we know we know that triangle C de C D E is similar to triangle to triangle C be a C be a and just from that you can get some interesting results because then we know that the ratio of this side of the shorter trial of the smaller triangle to the longer triangle is also going to be one half because the other two sides have a ratio of one half and we're dealing with similar triangles so this is going to be one half of that and we know one half of a B is just going to be is just going to be the length of F a 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 de over B a over B a 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 let's think about this triangle up here let's think about this triangle up here triangle we could call it B D F triangle B D F so first of all if you compare triangle B D F 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 we can you can just look at this diagram and you know that the ratio of BA the ratio let me do this way the ratio of bf2 ba the ratio of B F to B a is equal to 1/2 which is also the ratio of B D to BC B D 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 half of this whole length B F is half 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 by SAS similarity SAS similarity we know that triangle we know that triangle I'll write it this way d D B F triangle D b F is similar to triangle C B a is similar to triangle C B a 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 F D the ratio of F D to AC has to be 1/2 or F D has to be 1/2 of AC which when 1/2 of AC is just the length of a e 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 the corresponding angles have to be the same and we know that this that 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 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 a F over a B which is going to be the same as the ratio of a e over AC 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 the 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 we know that triangle EF a triangle e F a is similar to triangle C be a two triangle C be a and so the ratio all of the corresponding sides need to be one half so the ratio of Fe to BC needs to be one half R Fe needs to be half of that which is just the length of BD so this is just going to be that length right over there now and you can also say that since we've shown that all that this triangle this triangle and this triangle we haven't talked about this middle one yet they are 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 if the larger triangle had this blue angle right over here then 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 at this vertex right over here because this corresponds to that vertex based on the similarity so that's interesting now let's look at the let's look at 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 out of this one mark side this to mark side of this three mark side one mark to mark three mark one mark to mark three mark and that even applies to this middle triangle right over here so by side-side-side so by by side-side-side congruence congruence we now know and we want to be careful to get a corresponding sides right we now know that triangle triangle c de is congruent to triangle D to D BF d bf 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 to triangle e F a e F a which is going to be congruent to this triangle in here but we want to make sure that we're getting 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 we know and this is interesting that the the because the 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 in 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 blue angle sorry 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 congruence 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 F ed2 triangle f e d 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 the large they're all similar to the larger triangle - 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 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 corresponding angles here and here you could say that this is going to be parallel to that right over there and then finally make the same argument over here you have you have want to make sure I get the right corresponding angles you have this this line and this line and this angle corresponds to that angle they're the same so this de must be parallel to be a to be a so that's another neat property of this medial triangle I thought did all of these things just jump out when you just try to do something fairly simple with a with a triangle