WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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For this next lesson, we are going to start going over parallel lines and proportional parts.
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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.
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If a line is parallel to one side of a triangle, and it intersects the other two sides in two distinct parts,
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then it separates these sides into proportional segments.
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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;
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and that segment is parallel to a side; then these parts, these segments that are cut up, are going to be proportional.
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That means that AD over DB, that ratio, is going to be equal to AE over EC.
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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.
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No matter how you say it, no matter how you are going to state your scale factor, your ratios,
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make sure that you name the corresponding parts in that order: BD/DA is equal to CE/CA.
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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.
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That is triangle proportionality; now, this is the converse of that.
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If I have a triangle and a line intersects the two sides of a triangle at any point on these two sides
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so that these parts are now proportional, so AB/BC is equal to AE/ED, if this is true,
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then the line is parallel to the third side--they are parallel.
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The triangle proportionality theorem stated that, if the lines are parallel, then the parts are proportional.
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This one is saying that, if the parts are proportional, then the lines are parallel; so it is just the converse.
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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.
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So, it means that these endpoints are now at the midpoints of the sides.
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A segment whose endpoints are the midpoints--that means that this point right here is the midpoint of CE, so then these are congruent;
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and point D is the midpoint of AC, so then those are congruent.
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Then, in that case, well, this still applies the triangle proportionality theorem, where it is going to be parallel,
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and its length is one-half the length of the third side.
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Let me just do this different color; that way, you can see the conclusion to this.
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DE is parallel to AB, and DE is half the measure of AB.
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If you remember, when we went over quadrilaterals, we went over the trapezoid, and we went over the mid-segment.
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It is almost the same thing; the only difference is that, when we have the trapezoid,
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the mid-segment, which is the midpoint of the two sides, was half of the two bases added together--
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the sum of the two bases, divided by two, was the mid-segment, because there were two bases.
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When there are two bases, you have to add them up, and then you multiply by one-half.
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In this case, it is the same concept, the same idea; but it is just that we only have one base.
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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.
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If you remember this, it is the same concept--it is just that you can apply it here, too.
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Again, when you have that segment in the triangle whose endpoints are the midpoints of the sides, then two things:
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it is parallel (and that is just from the triangle proportionality theorem), and it is half the measure of the base, AB.
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The next one: parallel lines and transversals: now, with this, it is just saying that, if you have three or more parallel lines--
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these are all parallel (parallel, parallel, and parallel) lines, and there are two transversals here--
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now, when they intersect the two transversals (it doesn't matter how the transversals look), then they cut off the transversals proportionally.
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That means that this part to this part is going to be equal to this part to this part.
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So then, the two parts of the transversals are now proportional, only if those lines are parallel.
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If they are not parallel, then none of this applies.
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They have to be parallel, and then now, these parts are proportional.
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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.
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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.
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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.
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If we have congruent segments here on one transversal (they are going to cut off congruent segments),
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then it is going to do the exact same thing to every other transversal that it is cutting off.
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It is just that these two say the same thing: whatever the parallel lines do to one transversal, they do to every transversal.
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And then, if there are two of them, then their parts are now proportional.
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Let's work on our examples: Complete each statement.
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If we have IA/IC, that is going to be equal to something over ID.
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Now, before we continue that one, let's take a look at this diagram.
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Here is my triangle here, ACD; I have these lines that are intersecting the triangle, and they are all parallel.
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And there is this transversal that is cutting these parallel lines, and this one, also.
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The triangle proportionality theorem, remember, says that, if you have a segment that is touching the two sides
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where it is parallel (so that applies to this and to this, because they are both parallel to this side),
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then their parts are proportional; so then, all of these would be proportional.
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And then, this transversal--with that one, remember, it is saying that here, with these parallel lines,
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this could be considered a transversal, and this could be considered a transversal, because these lines are cutting them off here--
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cutting this transversal, and this one, and this one; that would be like three transversals, right there.
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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.
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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.
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The next one: GD, this part right here, over something, is going to be equal to HG over FE.
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So, this ratio is going to be equal to this to this--GD to what part?
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It is ED, because we are talking about this transversal and this transversal.
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GD/ED is going to equal HG/FE, so here, it is going to be ED.
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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,
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because they are all parallel, and here we have the transversals, and here we have the triangle, for triangle proportionality.
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The next example: Determine if the statement is true or false.
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BC...if this segment right here is 3/5 of AC, the whole thing, then AE to ED is 3/2.
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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,
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BC to AC; then that will help me with this side, with AE and ED.
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If I take this whole thing right here, BC, and make it equal to 3/5 of AC,
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well, if I take this right here, this is 3/5 times AC.
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So, what I can do is just divide the AC to both sides; then BC/AC is equal to 3/5.
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Then, doesn't that mean that, because fractions are ratios, too, then BC to AC is 3/5?
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That means that, if the ratio of this is 3, then the whole thing has to be 5.
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So then, what would this be? 2.
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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.
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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.
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So then, these are just the ratios of them.
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Well, since I know, from the triangle proportionality theorem, that, if I have a triangle,
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and its segment is intersecting the two sides, and it is parallel, then these parts are now proportional to these parts.
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If this has a ratio of 2:3, then this has to have a ratio of 2:3.
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So then again, I am going to write the ratios here; and this whole thing has to be 5.
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AE has to be 2, to ED, which is 3; the ratio has to be 2:3.
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It doesn't say 2:3; it says 3:2; so then, this one would be wrong--this one is false.
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If this is 2:3, then this cannot be 3:2; it has to have the same ratio.
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The next one: Find the values of x and y.
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I don't see that any of the sides are parallel--no parallel segments.
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All we have is that these are congruent and these parts are congruent.
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That means that this segment right here is considered the mid-segment,
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because this endpoint is at the midpoint of this side, and this endpoint is at the midpoint of this side.
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That means, remember, two things: that automatically, if those endpoints are the midpoints,
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then these are parallel; so then, these sides would be parallel; and this side is going to be half the length of this side.
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So, x - 3, the mid-segment, is going to be half the length of that base, 6/5x + 10.
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And then, from here, I can solve for x: x - 3 equals...make sure you distribute that 1/2; this becomes 3/5x + 5.
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I am going to add the 3; x = 3/5x + 8; and then, if you subtract this from here, then this is 1x;
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so then, this is 5/5x; that is the same thing, because 5/5 is 1;
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and the reason why I am making it 5/5 is because I need a common denominator to be able to subtract those fractions.
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I am just going to write 5/5x, which is the same thing as 1x, which is the same thing as x.
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So here, this becomes 2/5x, which is equal to 8.
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Then, how do I solve for x? Well, to get rid of this, I can just multiply it by its reciprocal.
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Then, whatever I do to that side, I have to do to this side.
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That way, this will cross-cancel out, and this just becomes x.
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And then, this becomes 1; this becomes 4; so x is 20.
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All of this here is just algebra; it is just being able to come up with this equation right here;
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and then, from there, you just distribute; then we just subtract this x over to there,
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and then we add the 3 over to the other side, and we just solve for x.
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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.
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So, 2/5x is equal to 8, because that went away; and then from there, I have to get rid of 2/5,
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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.
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And then, x becomes 20; that is the value of x.
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And then, for y, y is going to be a little bit easier, because we know that,
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since this is the midpoint, this part and this part are congruent, and that shows it right there.
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So, since they are congruent, we could just make them equal to each other: 5y + 4 = 3y + 12.
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Subtract this y over; that gives me 2y; if I subtract the 4 over, it becomes 8.
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And then, y = 4; so those are my values--there is x, and there is y.
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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.
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If it asks for the actual lengths, then you would have to go back, plug it in, and solve for it.
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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.
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DE, this whole thing, is 20; DA, this, is 12; FC is 9.
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We are going to find AB, BC, and AC; we are going to find these three lengths.
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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.
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Now, let's see: first of all, let's look at one of these mid-segments at a time.
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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.
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That means that AC is half the length of DE; it is parallel, and it is half the length.
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If this whole thing is 20, I know that AC, this right here, is 10.
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I will write that in red, so that we will know that that is what we found.
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And what else? AC is 10, and then, we are just going to do that for each of them.
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Here, let's look at AB; AB is also a mid-segment, because this endpoint is the midpoint of this side;
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this endpoint is the middle of this side; therefore, this is parallel to the third side, and it is half the length.
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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.
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Half of 18 is each of these...9...that would also make this 9.
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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.
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And then, BC, again, has a midpoint at that side, and a midpoint at that side;
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this side, BC, is parallel to DF, and it is half the length; so half the length, we know, is 12,
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because this is half the length, and the whole thing is going to be 24; so this is going to be 12.
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For this example, it is just three mid-segments in one triangle; so here is AB, BC, and AC.
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And that is it for this lesson; thank you for watching Educator.com.