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### Parts of Similar Triangles

- Proportional perimeters: If two triangles are similar, then the perimeters are proportional to the measure of corresponding sides
- Similar Altitudes: If two triangles are similar, then the measure of the corresponding altitudes are proportional to the measure of the corresponding sides
- Similar Angle Bisectors: If two triangles are similar, then the measures of the corresponding angle bisectors of the triangles are proportional to the measures of the corresponding sides
- Similar Medians: If two triangles are similar, then the measures of the corresponding medians are proportional to the measures of the corresponding sides
- Angle Bisector Theorem: An angle bisector in a triangle separates the opposite side into segments that have the same ratio as the other two sides

### Parts of Similar Triangles

If two triangles are similar, then the measures of the corresponding altitudes, perimeters, medians and angle bisectors are all proportional to the measures of the corresponding sides.

m∠BAD = m∠CAD, complete the statement.

[AB/AC] = [BD/x]

∆ABC:∆DEF, AB = 12, DE = 8, perimeter of ∆ABC = 36, find the perimeter of ∆DEF

- [(perimeter of ∆ ABC)/(perimeter of ∆ DEF)] = [AB/DE]
- [36/(perimeter of ∆ DEF)] = [12/8]
- perimeter of ∆DEF = 36*8 ÷12

∆ABC ∼ ∆DEF, ―AP and ―DQ are medians of the two triangles

AB = 12, DE = 8, AP = 2x + 2, DQ = x + 2, find x.

- [AP/DQ] = [AB/DE]
- [(2x + 2)/(x + 2)] = [12/8]
- 16x + 16 = 12x + 24

An angle bisector in a triangle separates the opposite side into segments that have the same ratio as the other two sides.

If ∆ABC ∼ ∆DEF, ―AP and ―DQ are the bisectors of the two triangles, then [AB/DE] = [AP/DQ].

If ―AP is the median of ∆ABC, then [BP/CP] = [AB/AC].

∆ABC ∼ ∆DEF, ―AM and ―DN are altitudes of the two triangles

AB = 16, DE = 8, AM = 3x + 8, DN = x + 6, find x.

- [AM/DN] = [AB/DE]
- [(3x + 8)/(x + 6)] = [16/8] = 2
- 3x + 8 = 2x + 12

An altitude in a triangle separates the opposite side into segments that have the same ratio as the other two sides.

∆ABC ∼ ∆DEF, ―BP and ―EQ are medians of the two triangles

AB = 16, DE = 8, BP = 12, find EQ.

- [AB/DE] = [BP/EQ]
- [16/8] = [12/EQ]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Parts of Similar Triangles

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Proportional Perimeters 0:09
- Proportional Perimeters: Definition and Example
- Similar Altitudes 2:23
- Similar Altitudes: Definition and Example
- Similar Angle Bisectors 4:50
- Similar Angle Bisectors: Definition and Example
- Similar Medians 6:05
- Similar Medians: Definition and Example
- Angle Bisector Theorem 7:33
- Angle Bisector Theorem
- Extra Example 1: Parts of Similar Triangles 10:52
- Extra Example 2: Parts of Similar Triangles 14:57
- Extra Example 3: Parts of Similar Triangles 19:27
- Extra Example 4: Find the Perimeter of Triangle ABC 23:14

### Geometry Online Course

### Transcription: Parts of Similar Triangles

*Welcome back to Educator.com.*0000

*Now that we have gone over similar triangles, we are going to go over parts of those similar triangles for this lesson.*0003

*Let's talk about their perimeters: now, we know that the perimeter is the sum of all of the sides.*0013

*When it comes to these triangles, if they are similar, then their perimeters are going to have the same scale factor, the same ratio, as these corresponding sides.*0020

*So, if, let's say, AB has a side of 8, and DE, the corresponding side, has a measure of 10;*0042

*then the scale factor between these two triangles would be 8/10; or to simplify, it would be 4/5.*0056

*So, this is the scale factor; that is the ratio.*0066

*And then, since they are similar, we now know that their perimeters are going to have the same scale factor.*0070

*The perimeters are going to be proportional to each other.*0077

*The perimeter for this triangle--let's say it is 20: if the perimeter here is 20, then the perimeter here we can solve for, using the same ratio.*0081

*So, if this is 4:5, then remember: I can make a proportion so that it will have the same ratio: 20/x.*0095

*We do cross-products: 4x = 100; divide the 4; x is 25, so the perimeter of this triangle is 20, and the perimeter of this triangle is going to be 25.*0109

*So again, as long as the triangles are similar, then the perimeters are going to have the same scale factor,*0128

*or they are going to be proportionally the same as the corresponding sides.*0137

*Now, in the same way, altitudes, remember, are segments that start from the vertex (or whose one endpoint is on the vertex),*0145

*and they go down so that the other endpoint is going to be on one of the sides, so that the segment is perpendicular to that side.*0159

*If you have a segment like that, and it is perpendicular, that is called an altitude.*0173

*If you have an altitude drawn from a triangle, and an altitude from another triangle, and those two triangles are similar,*0180

*then those altitudes are also going to have the same ratio, or they are also going to be equally proportional.*0192

*They are going to be just the same; so here, if I give this point...say I label that P, and then this one I label N,*0199

*then this altitude, BP, to the other altitude, EN, is going to have the same ratio, or scale factor, as any of the other sides.*0211

*We already know that corresponding sides are going to be proportional, as long as the triangles are similar.*0226

*So, we already know that AB to DE...that ratio is going to be the same as AC to DF and BC to EF.*0231

*So then, they are all proportional; now we just say that those altitudes are going to be also proportional; they are going to have the same scale factor.*0240

*So then, I can say BP/EN is going to be equal to...and then, any pair of corresponding sides, because it is going to be all the same.*0249

*I can say that, since I named this altitude first, I have to name this triangle, so it is AB/DE.*0258

*And again, for this one, I can name any corresponding sides; so BP/EN is equal to BC/EF, or AC/DF, and so on.*0270

*That is altitude: a segment so that it is perpendicular to the side.*0282

*And angle bisectors: an angle bisector is like an altitude, but then, it is not perpendicular.*0292

*An angle bisector, remember, is something that cuts, in this case, an angle, in half; it bisects the angle.*0299

*So again, it is starting from the vertex, going to the side opposite that vertex, so that the angle is bisected; that is the angle bisector.*0310

*So again, if we have two triangles that are similar, then the angle bisectors are going to be proportional*0322

*with the same scale factor as the corresponding sides of any one of those from the triangles.*0337

*So, I can say that BN/EP is going to have the same scale factor as AB/DE.*0344

*And the last one, the median--with the median also, the endpoints are on the vertex and the side opposite that vertex,*0367

*so that it cuts that side into two equal parts, bisecting the side; that is the median.*0376

*BN is a median of this triangle; EP is a median of this triangle; these two triangles are similar, so then,*0384

*these medians are going to be proportional, with the same scale factor as any corresponding sides from those triangles.*0394

*BN to EP, the other median, is going to be the same as AB to DE, or I can say equal to BC to EF, which is also equal to AC to DF.*0406

*Now, make sure that, if you are going to name BN first, BN to EP, then for the second ratio, you have to name a part from this triangle first.*0428

*So, you can't say BN to EP equals DE to AB; you can't say that--it has to be in the same order.*0441

*Now, this one is a little bit different; it is not saying that the medians or the angle bisectors*0455

*or the altitudes or the perimeters are all going to have the same proportional measure as the sides.*0461

*This one is talking about the angle bisector; it is the angle bisector theorem where, if I have this angle bisector, CD,*0469

*then this angle bisector is going to cut this triangle into proportional segments.*0483

*It just means that, from here, I have AD and DB.*0494

*So now, these parts are going to be proportional to the other two sides.*0501

*To just read it, an angle bisector in a triangle separates the opposite side (which is this right here)*0510

*into segments that have the same ratio as the other two sides.*0518

*These two parts right here...AD over DB is going to be the same ratio as the other two sides, AC over BC.*0525

*Again, AD, this part, to DB, equals AC to BC.*0543

*This was a little bit different than what we just went over for angle bisectors,*0551

*because that is saying, "Well, if I have two triangles and an angle bisector, then this angle bisector*0555

*with this angle bisector is going to have the same ratio as the other sides,"*0568

*because if you have two similar triangles, you have a scale factor between them,*0574

*meaning the ratio between this triangle and this triangle.*0580

*And it is just saying that those two different angle bisectors are going to have that same ratio between them.*0584

*This one is different, because it is only giving you one angle bisector.*0591

*It is not comparing an angle bisector with an angle bisector from another triangle.*0595

*This one is one single angle bisector, and that one angle bisector is going to cut this one triangle up into different segments that are going to be proportional.*0600

*So again, AD to DB: those segments are going to be proportional to AC to BC, the other two sides.*0614

*That is the angle bisector theorem: there are two different theorems that we have gone over for the angle bisector.*0632

*The first one: it is talking about two different angle bisectors; this one is one single bisector*0639

*that is cutting this triangle up into different parts that are going to have the same ratio.*0645

*Let's go over some examples: the first one...we are just going to use these two; this is actually #2 (let me write that in).*0655

*The first one: there are two triangles, and they are similar.*0664

*And this right here--those are medians, because they are bisecting the side right here.*0672

*So then, there are two equal parts (segments); that is the median.*0683

*And it doesn't matter, because they are all the same; it is just different theorems, but then, it is all the same concept--*0689

*that they are all going to have the same ratio.*0695

*I can say that 15, the first median, over...the second median is x...equals one side (that is 10), over another side (is 18).*0700

*These are corresponding sides, so we can use that ratio.*0717

*From here, now, I am going to do cross-products.*0721

*But see here: 15 times 18--that is kind of a big number.*0726

*I am going to simplify this; this is the same as 5/9.*0731

*I can just use this; this is equal to this equivalent fraction; I am just going to use this.*0740

*And it is just easier to just multiply 15 times 9.*0746

*You can go ahead and use that; it is fine to multiply that; but I am just going to make it 5/9.*0749

*Then, 15 times 9 is 135; 135 is equal to (that was this side right here) 5x.*0757

*So, divide the 5, and you get 27; so that is equal to x.*0776

*And then, #2: here, we have altitudes, this altitude and this altitude.*0791

*And we know that it is an altitude because it is perpendicular to the third side.*0804

*This one is not perpendicular; it is at the midpoint of the sides; so here, we have altitudes (let me just write that in).*0808

*So then, this one here that we had...these are medians...and then, for this one, we have altitudes.*0816

*Again, we are going to use the same strategy by doing altitude over altitude, equal to side over side.*0832

*And then, here, I can also simplify this to make that 3/4; 3/4 = 12/x; 4 times 12 is 48; that is equal to 3x.*0844

*Divide the 3; then I get 19 = x...no, that isn't 19; 16 = x.*0861

*Here, we know that this median is going to be 27; and here, we know that this side is going to be 16.*0886

*OK, the next one: Find the value of DC.*0896

*There are a few things to look at here: We have an angle bisector, because of this; this shows me that it is bisecting this angle.*0903

*And then, let's see, I don't have two similar triangles; I have that this side is 18; this little part right here is 4; BC is 14.*0915

*I know that this is also 14, because these are congruent.*0930

*I don't know what this side is, or this side; and this is the side I am looking for.*0933

*So, let's see: what can we use here?*0940

*We know that the angle bisector theorem says that, if we have an angle bisector...*0944

*now, this angle bisector can be for this triangle, or it could be for the big triangle; so either way, here,*0951

*it is going to be part to part equals part to part (the other two sides), or the angle bisector if I am looking at the big triangle--*0961

*then it is going to be part to part equals side to side.*0972

*But then, I don't really know what these parts are; I can't say that this is half exactly, because I don't know.*0980

*This is not a median; if this was a median, then I know that this would be half of that measure, but it is not.*0988

*Here, these two sides being congruent, and these two parts being congruent for this side--*0998

*that gives me more information; that kind of helps me out.*1006

*From the previous lesson, we know that, if I have a segment within a triangle,*1010

*if I look at the big triangle, and then I think of this as a segment, and this endpoint is the midpoint of this side,*1020

*and this endpoint is the midpoint of this side, then I know that this is the mid-segment.*1029

*It is the middle segment so that this is parallel to EA, and it is half the measure of EA.*1037

*Remember that? If DB is half the measure of EA, well, EA is 18; that means this whole thing right here is going to be 9.*1046

*This DF is 4; what would that make FB? That would make that 5.*1062

*Now, I am going to label that x; so, now I know that I can use my angle bisector theorem,*1070

*because right here, if I look at the smaller triangle, I have this part; I know that part over part is going to equal side over side.*1081

*And I have these parts, and I have this side, so only this side is missing.*1094

*That means that I can go ahead and write my proportion: 4/5 (remember, part to part) is going to equal DC, or x, over BC, which is 14.*1099

*So, from here, I am going to do cross-products: 14 times 4 is going to be 56; that equals 5 times DC.*1118

*I am going to divide the 5, and then I know that DC is going to be 56/5.*1138

*And you can just go ahead and leave it like that; so DC is 56/5.*1146

*Or if you want to, you can just make it into a mixed number, which is 11 and 1/5, or 11.2.*1151

*But then, that should be fine: DC is 56/5.*1161

*The next one: BD and NP (here is BD, and here is NP) are medians.*1167

*That means...well, they didn't show you that they are medians; they actually told you that they were medians.*1183

*So then, you can just draw in these little slash marks, and then find the value of x.*1188

*We know that medians are going to have the same ratio as the ratio between similar triangles.*1197

*I can say that 11, the first median, over the second median, 10, is going to equal 2x - 1, over 3x - 4.*1212

*Now, you are probably thinking, "Well, why is this ratio this part and this part--it is not the whole side?"*1230

*The theorem before on the median said that median to median is equal to side to side.*1240

*But this is not a whole side; it is just part of a side.*1248

*But it would actually be the same thing, because they are medians; and let me just explain that.*1252

*If this part is 2x - 1, this whole thing is going to be 2 times 2x - 1, because this is 2x - 1, and this is also 2x - 1.*1258

*So, it is 2 times 2x - 1; well, this whole thing is going to be 2 times 3x - 4.*1272

*So, if you want to use a theorem and say that 11, the median, over median, is equal to side,*1281

*which is 2(2x - 1), over side, 2(3x - 4), what happens here?*1289

*These 2's will cancel out, so it just becomes the same thing as that.*1300

*And it is only like that, because they are medians; and then, those two sides are equal.*1306

*From here, let's go ahead and solve it; cross-multiply; 11 times all that...let me just write it out.*1314

*Then, I can distribute, so that becomes 33x - 44 = 20x - 10.*1330

*If I subtract the 20x, that becomes 13x =...I am going to add the 44, so I get 34.*1342

*And then, I divide the 13, so x is going to equal 34/13.*1355

*I know that this is not going to be able to simplify; they don't have a common factor; so that is my answer, 34/13.*1364

*Again, you can just leave it like that, or you can make it a decimal, or you can make it into a mixed number, if you would like.*1375

*But that is the answer for x; so again, just make it into a proportion; just go ahead and cross-multiply and solve it out.*1385

*The fourth example: Find the perimeter of triangle ABC.*1395

*Here, all you are given is the length of AB and DE; and again, we are going to say that this is similar.*1401

*The perimeter is triangle DEF is given; and then, we are going to look for the perimeter here.*1410

*So then, the perimeter is what we are solving for.*1415

*Remember the first theorem that we went over for this lesson; it said that the perimeters are going to have the same ratios as the triangles' corresponding sides.*1423

*So, the ratio from AB to DE is going to be 10.2/12.5.*1434

*It is going to equal...I am just going to write P for perimeter, and that is this triangle here...over 32, the perimeter of that one.*1447

*OK, so then, you are going to multiply this.*1458

*If you multiply...let's do that over here, and I will just multiply it out...this becomes 402; this becomes 603, 326.4...*1460

*OK, let me just double-check: 402, 603, and then my decimal goes right there: 326...if you have a calculator,*1479

*you can just use your calculator...that equals 12.5P.*1492

*And then, you have to divide the 12.5, so it is all that divided by 12.5.*1499

*And then, for this one, you wouldn't be able to leave it like that, because you have decimals within your fraction.*1509

*When you have decimals within your fraction, you want to simplify.*1516

*You want to either change your whole fraction to a decimal, or just simplify so that your decimals go away.*1519

*326.4, divided by 12.5: this is going to be...and then, you can just go ahead and use your calculator.*1528

*That is the easiest way to do this; for us here, we are going to say 2...this becomes 250;*1548

*if I subtract that, it is going to become 76, with a 4; and then, let's see, 250, 500, 750...so then, let's say 6.*1560

*We multiply this by 6, and I am going to get 30, 12, 15, 7; so then, 6...and then, here is the 0...and that is .1.*1582

*So, I am just going to leave it like that...something around there.*1601

*The perimeter is 26.1, and then you could just use your calculator for that; that is the perimeter.*1608

*That is it for this lesson; thank you for watching Educator.com.*1623

0 answers

Post by Taylor Wright on June 13, 2013

Why can't you solve with:

CA/EA = CD/DB ?

1 answer

Last reply by: Malayna Canfield

Wed Aug 7, 2013 1:15 PM

Post by Taylor Wright on June 13, 2013

At 5:51

Would the two angles formed by the bisector in both triangles all be equal, since these triangles are similar and the bisector is dividing them in two?

0 answers

Post by QIFAN YE on December 11, 2012

about extra example II, how comes triangle DFC ~ triangle BFC?