Geometry Parallel Lines and Proportional Parts

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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 sides0145

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 sides0587

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