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 0 answersPost by Taylor Wright on June 13, 2013 Why can't you solve with:CA/EA = CD/DB ? 1 answerLast reply by: Malayna CanfieldWed Aug 7, 2013 1:15 PMPost by Taylor Wright on June 13, 2013At 5:51Would 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 answersPost by QIFAN YE on December 11, 2012about extra example II, how comes triangle DFC ~ triangle BFC?

### 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

Determine whether the following statement is true or false.
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.
True

[AB/AC] = [BD/x]
x = CD

∆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
perimeter of ∆DEF = 24

∆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
x = 2
Determine whether the following statement is true or false.
An angle bisector in a triangle separates the opposite side into segments that have the same ratio as the other two sides.
True
Determine whether the following statement is true or false.

If ∆ABC  ∼  ∆DEF, AP and DQ are the bisectors of the two triangles, then [AB/DE] = [AP/DQ].
True
Determine whether the following statement is true or false.

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

∆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
x = 4
Determine whether the following statement is true or false.
An altitude in a triangle separates the opposite side into segments that have the same ratio as the other two sides.
False

∆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]
EQ = 6

*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.

### 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

### 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 proportional0322

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 bisectors0455

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 bisector0555

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 bisector0639

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

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