### Parallel Lines and Proportional Parts

- If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments
- Triangle Proportionality Converse: If a line intersects two sides of a triangle and separates the sides into corresponding segments of proportional lengths, then the line is parallel to the third side
- Triangle Mid-segment: A segment whose endpoints are the midpoints of two sides of a triangle is parallel to the third side of the triangle, and its length is one-half the length of the third side
- If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally
- If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments one very transversal

### Parallel Lines and Proportional Parts

Determine whether the following statement is true or false.

If ―DE ||―BC , then [AD/DB] = [AE/EC].

Determine whether the following statement is true or false.

If D and E are the midpoints of ―AB and ―AC , then ―DE ||―BC .

Trapezoid ABCD,―MN ||―BC .

Complete the statement.

[AM/MB] = [AO/x]

Complete the statement.

[CO/AO] = [x/DN]

Trapezoid ABCD, ―MN ||―BC .

AO = 2x + 1, OC = 3, AM = 4, MB = 6, find x.

- [AM/MB] = [AO/OC]
- [4/6] = [(2x + 1)/3]
- 12x + 6 = 12

If a line intersects two sides of a triangle and separates the sides into corresponding segments of proportional lengths, the the line is _____parallel to the third side.

If the endpoints of a segment are the midpoints of two sides of a triangle, then the length of the segment is one - half the length of the third side.

D and E are the midpoints of ―AB and ―AC , DE = 2x + 3, BC = 3x + 8, find x.

- BC = 2DE
- 3x + 8 = 2(2x + 3)
- 3x + 8 = 4x + 6

Determine whether the following statement is true or false.

If ―DE ||―BC , then ∆ADE ∼ ∆ABC.

Trapezoid ABCD, E is the midpoint of ―AB

―EF ||―BC

AC = 10, BC = 14, EF = 12, find AG and FG.

- [AE/EB] = 1
- [AG/GC] = 1
- AG = GC = [1/2]AC
- AG = [1/2]*10 = 5
- EG = [1/2]BC = [1/2]*14 = 7
- FG = EF − EG
- FG = 12 − 7 = 5

*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

### Parallel Lines and Proportional Parts

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
- Triangle Proportionality 0:07
- Definition of Triangle Proportionality
- Example of Triangle Proportionality
- Triangle Proportionality Converse 2:19
- Triangle Proportionality Converse
- Triangle Mid-segment 3:42
- Triangle Mid-segment: Definition and Example
- Parallel Lines and Transversal 6:51
- Parallel Lines and Transversal
- Extra Example 1: Complete Each Statement 8:59
- Extra Example 2: Determine if the Statement is True or False 12:28
- Extra Example 3: Find the Value of x and y 15:35
- Extra Example 4: Find Midpoints of a Triangle 20:43

### Geometry Online Course

### Transcription: Parallel Lines and Proportional Parts

*Welcome back to Educator.com.*0000

*For this next lesson, we are going to start going over parallel lines and proportional parts.*0002

*Triangle proportionality: here we have a triangle with a line segment whose endpoints are on the side of the triangle, and it is parallel to a side.*0010

*If a line is parallel to one side of a triangle, and it intersects the other two sides in two distinct parts,*0028

*then it separates these sides into proportional segments.*0041

*That means that, if I have a triangle, and the segment is in the triangle, the endpoints have to be touching the sides of the triangle;*0052

*and that segment is parallel to a side; then these parts, these segments that are cut up, are going to be proportional.*0064

*That means that AD over DB, that ratio, is going to be equal to AE over EC.*0081

*You don't have to say it this way; you can also say this part first: you can say that BD/DA is equal to CE/EA.*0094

*No matter how you say it, no matter how you are going to state your scale factor, your ratios,*0110

*make sure that you name the corresponding parts in that order: BD/DA is equal to CE/CA.*0115

*Or if you are going to mention it like this, with AD first, then for the second ratio, you have to mention the corresponding part (AE) first.*0125

*That is triangle proportionality; now, this is the converse of that.*0135

*If I have a triangle and a line intersects the two sides of a triangle at any point on these two sides*0145

*so that these parts are now proportional, so AB/BC is equal to AE/ED, if this is true,*0163

*then the line is parallel to the third side--they are parallel.*0186

*The triangle proportionality theorem stated that, if the lines are parallel, then the parts are proportional.*0202

*This one is saying that, if the parts are proportional, then the lines are parallel; so it is just the converse.*0213

*The triangle mid-segment: this is the same thing, but it is now saying that this segment right here is right between this side and that vertex.*0223

*So, it means that these endpoints are now at the midpoints of the sides.*0241

*A segment whose endpoints are the midpoints--that means that this point right here is the midpoint of CE, so then these are congruent;*0249

*and point D is the midpoint of AC, so then those are congruent.*0259

*Then, in that case, well, this still applies the triangle proportionality theorem, where it is going to be parallel,*0266

*and its length is one-half the length of the third side.*0280

*Let me just do this different color; that way, you can see the conclusion to this.*0285

*DE is parallel to AB, and DE is half the measure of AB.*0293

*If you remember, when we went over quadrilaterals, we went over the trapezoid, and we went over the mid-segment.*0315

*It is almost the same thing; the only difference is that, when we have the trapezoid,*0328

*the mid-segment, which is the midpoint of the two sides, was half of the two bases added together--*0336

*the sum of the two bases, divided by two, was the mid-segment, because there were two bases.*0350

*When there are two bases, you have to add them up, and then you multiply by one-half.*0357

*In this case, it is the same concept, the same idea; but it is just that we only have one base.*0362

*See how the other side is just a vertex; so it is just the base times one-half, but this one is the bases times one-half.*0366

*If you remember this, it is the same concept--it is just that you can apply it here, too.*0380

*Again, when you have that segment in the triangle whose endpoints are the midpoints of the sides, then two things:*0388

*it is parallel (and that is just from the triangle proportionality theorem), and it is half the measure of the base, AB.*0398

*The next one: parallel lines and transversals: now, with this, it is just saying that, if you have three or more parallel lines--*0414

*these are all parallel (parallel, parallel, and parallel) lines, and there are two transversals here--*0422

*now, when they intersect the two transversals (it doesn't matter how the transversals look), then they cut off the transversals proportionally.*0432

*That means that this part to this part is going to be equal to this part to this part.*0441

*So then, the two parts of the transversals are now proportional, only if those lines are parallel.*0452

*If they are not parallel, then none of this applies.*0462

*They have to be parallel, and then now, these parts are proportional.*0465

*Let's say that this is 2, and this is 3; let's say that this is 4, and this has to be 6, because they are now proportional: 2/3, 4/6.*0472

*This second statement: if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.*0490

*That just means that, if we have these parallel lines, whatever these lines do to one transversal, it is going to do the exact same thing to another transversal.*0501

*If we have congruent segments here on one transversal (they are going to cut off congruent segments),*0511

*then it is going to do the exact same thing to every other transversal that it is cutting off.*0515

*It is just that these two say the same thing: whatever the parallel lines do to one transversal, they do to every transversal.*0522

*And then, if there are two of them, then their parts are now proportional.*0531

*Let's work on our examples: Complete each statement.*0541

*If we have IA/IC, that is going to be equal to something over ID.*0545

*Now, before we continue that one, let's take a look at this diagram.*0556

*Here is my triangle here, ACD; I have these lines that are intersecting the triangle, and they are all parallel.*0564

*And there is this transversal that is cutting these parallel lines, and this one, also.*0576

*The triangle proportionality theorem, remember, says that, if you have a segment that is touching the two sides*0587

*where it is parallel (so that applies to this and to this, because they are both parallel to this side),*0596

*then their parts are proportional; so then, all of these would be proportional.*0602

*And then, this transversal--with that one, remember, it is saying that here, with these parallel lines,*0611

*this could be considered a transversal, and this could be considered a transversal, because these lines are cutting them off here--*0621

*cutting this transversal, and this one, and this one; that would be like three transversals, right there.*0627

*Back to the first one: you have IA, this part, over the whole thing, IC; that is going to be equal to something over ID, this whole thing.*0639

*That is going to be IF; it has to be IF; this part over the whole is going to be equal to this part over the whole; x is going to be IF.*0659

*The next one: GD, this part right here, over something, is going to be equal to HG over FE.*0674

*So, this ratio is going to be equal to this to this--GD to what part?*0689

*It is ED, because we are talking about this transversal and this transversal.*0701

*GD/ED is going to equal HG/FE, so here, it is going to be ED.*0708

*And again, they are just all the corresponding parts; there are so many parts, but just know that they are all going to be proportional,*0724

*because they are all parallel, and here we have the transversals, and here we have the triangle, for triangle proportionality.*0735

*The next example: Determine if the statement is true or false.*0749

*BC...if this segment right here is 3/5 of AC, the whole thing, then AE to ED is 3/2.*0754

*BC is 3/5 of AC; now, I want to know what the ratio is between BC and AC, because it is using those two segments,*0775

*BC to AC; then that will help me with this side, with AE and ED.*0792

*If I take this whole thing right here, BC, and make it equal to 3/5 of AC,*0805

*well, if I take this right here, this is 3/5 times AC.*0815

*So, what I can do is just divide the AC to both sides; then BC/AC is equal to 3/5.*0820

*Then, doesn't that mean that, because fractions are ratios, too, then BC to AC is 3/5?*0836

*That means that, if the ratio of this is 3, then the whole thing has to be 5.*0845

*So then, what would this be? 2.*0852

*Now, it doesn't mean that that is the actual length; it just means that the ratio between those three parts is 3:2, and then the whole thing to 5.*0855

*That means that if AB has a length of 4, then BC would have a length of 6; AC would have a length of 10.*0865

*So then, these are just the ratios of them.*0874

*Well, since I know, from the triangle proportionality theorem, that, if I have a triangle,*0877

*and its segment is intersecting the two sides, and it is parallel, then these parts are now proportional to these parts.*0883

*If this has a ratio of 2:3, then this has to have a ratio of 2:3.*0896

*So then again, I am going to write the ratios here; and this whole thing has to be 5.*0902

*AE has to be 2, to ED, which is 3; the ratio has to be 2:3.*0908

*It doesn't say 2:3; it says 3:2; so then, this one would be wrong--this one is false.*0918

*If this is 2:3, then this cannot be 3:2; it has to have the same ratio.*0928

*The next one: Find the values of x and y.*0935

*I don't see that any of the sides are parallel--no parallel segments.*0943

*All we have is that these are congruent and these parts are congruent.*0951

*That means that this segment right here is considered the mid-segment,*0957

*because this endpoint is at the midpoint of this side, and this endpoint is at the midpoint of this side.*0963

*That means, remember, two things: that automatically, if those endpoints are the midpoints,*0976

*then these are parallel; so then, these sides would be parallel; and this side is going to be half the length of this side.*0985

*So, x - 3, the mid-segment, is going to be half the length of that base, 6/5x + 10.*1003

*And then, from here, I can solve for x: x - 3 equals...make sure you distribute that 1/2; this becomes 3/5x + 5.*1021

*I am going to add the 3; x = 3/5x + 8; and then, if you subtract this from here, then this is 1x;*1037

*so then, this is 5/5x; that is the same thing, because 5/5 is 1;*1058

*and the reason why I am making it 5/5 is because I need a common denominator to be able to subtract those fractions.*1066

*I am just going to write 5/5x, which is the same thing as 1x, which is the same thing as x.*1074

*So here, this becomes 2/5x, which is equal to 8.*1079

*Then, how do I solve for x? Well, to get rid of this, I can just multiply it by its reciprocal.*1089

*Then, whatever I do to that side, I have to do to this side.*1097

*That way, this will cross-cancel out, and this just becomes x.*1103

*And then, this becomes 1; this becomes 4; so x is 20.*1110

*All of this here is just algebra; it is just being able to come up with this equation right here;*1118

*and then, from there, you just distribute; then we just subtract this x over to there,*1128

*and then we add the 3 over to the other side, and we just solve for x.*1136

*Here, again, 5/5x is there because 5/5 is the same thing as 1, and you have to have a common denominator to subtract these fractions.*1141

*So, 2/5x is equal to 8, because that went away; and then from there, I have to get rid of 2/5,*1151

*this fraction; so I need to solve for x; you can also just divide this whole thing by 2/5, which would be the exact same thing.*1161

*And then, x becomes 20; that is the value of x.*1172

*And then, for y, y is going to be a little bit easier, because we know that,*1179

*since this is the midpoint, this part and this part are congruent, and that shows it right there.*1184

*So, since they are congruent, we could just make them equal to each other: 5y + 4 = 3y + 12.*1194

*Subtract this y over; that gives me 2y; if I subtract the 4 over, it becomes 8.*1207

*And then, y = 4; so those are my values--there is x, and there is y.*1217

*And go back to your directions; it just asks for the values of x and y, so then you can just leave those as the answer.*1227

*If it asks for the actual lengths, then you would have to go back, plug it in, and solve for it.*1235

*The last one: A, B, and C are each midpoints of the sides--they are midpoints of each of the sides of the triangle DEF.*1245

*DE, this whole thing, is 20; DA, this, is 12; FC is 9.*1258

*We are going to find AB, BC, and AC; we are going to find these three lengths.*1275

*And since they are the midpoints, I can just go ahead and do that, and then this one with this one, and this one to this one.*1285

*Now, let's see: first of all, let's look at one of these mid-segments at a time.*1297

*If we look at the segment AC, that is the mid-segment, because again, the endpoints are at the midpoints of each of the sides.*1309

*That means that AC is half the length of DE; it is parallel, and it is half the length.*1320

*If this whole thing is 20, I know that AC, this right here, is 10.*1325

*I will write that in red, so that we will know that that is what we found.*1334

*And what else? AC is 10, and then, we are just going to do that for each of them.*1339

*Here, let's look at AB; AB is also a mid-segment, because this endpoint is the midpoint of this side;*1354

*this endpoint is the middle of this side; therefore, this is parallel to the third side, and it is half the length.*1363

*Now, this right here is not half of 9, because, if this is 9, then this is also 9, which makes the whole thing 18.*1371

*Half of 18 is each of these...9...that would also make this 9.*1384

*It is just half of the whole thing; each of these is half of the whole thing also, so they are going to have the same measure.*1392

*And then, BC, again, has a midpoint at that side, and a midpoint at that side;*1401

*this side, BC, is parallel to DF, and it is half the length; so half the length, we know, is 12,*1411

*because this is half the length, and the whole thing is going to be 24; so this is going to be 12.*1417

*For this example, it is just three mid-segments in one triangle; so here is AB, BC, and AC.*1430

*And that is it for this lesson; thank you for watching Educator.com.*1441

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