WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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This next lesson is on special segments within triangles.
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The first one (there are a few) is the perpendicular bisector.
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Now, we know that "perpendicular" means that it is going to form a right angle.
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And a "bisector" is a segment or an angle that cuts a segment or an angle in half.
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So, a perpendicular bisector is going to be a line or a line segment that passes through the midpoint
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(that is the bisector part of it) and is perpendicular to that side.
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Two things: the segment has to pass through the side so that it is going to be perpendicular, and it is going to bisect that side.
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So, if I want to draw the perpendicular bisector of the side AC, I have to draw a line or a line segment
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that is going to be perpendicular to the side and is going to bisect it--it is going to cut it in half at its midpoint.
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Let's say that that is the midpoint right there; that means that this is cut in half.
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And then, I have to draw a segment that is going to be perpendicular and is going to bisect it--something like that,
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where it is perpendicular and it bisects; so this right here is going to be the perpendicular bisector.
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So again, the perpendicular bisector is a line or a line segment that is going to make it perpendicular and bisect the side.
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Now, I could draw a perpendicular bisector for each side of the triangle; this is just for side AC.
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But since I have three sides, I can draw three perpendicular bisectors: one for each side.
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If you are only required to draw one, then you can just draw it like this.
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You don't have to draw it all the way through; it can be a line or a line segment, so you can draw a line,
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or you can just cut it right here and just make it a segment.
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But that would just be for one of the sides; you can draw three, for each of the sides.
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If you wanted to draw a perpendicular bisector for side AB, it might be helpful for you to just turn your paper so that this is the horizontal side.
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So again, we want to have the midpoint, and then I am going to do that to show that that is the midpoint.
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And then, I need to draw a segment that is going to be perpendicular, like that.
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That is two; the third one is going to be for side BC.
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So again, ignore this line right here; so then, that would be 1, 2, 3, to show that those two parts are congruent.
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And then, draw...like that.
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What should happen is that all three perpendicular bisectors, the ones that you draw for each of the sides, should all meet at one point.
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Again, we have a perpendicular bisector; it is a line or line segment that is perpendicular to the side and bisects the side.
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That means that it cuts it in half; that is a perpendicular bisector.
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And then, for the perpendicular bisector, a couple of theorems: Any point on the perpendicular bisector of a segment...
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when you just draw another triangle, and let's say my perpendicular bisector is right there,
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then what this theorem is saying is that any point on the perpendicular bisector
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(now, this is the perpendicular bisector)--any point on this line is equidistant from the endpoints of the segment.
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So, this is the segment, AB; I could pick any point on this perpendicular bisector, and the distance from this point to this endpoint,
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and this point to this endpoint, is going to be the same.
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"Equidistant" means that it is the same distance; from this point to point A and this point to point B is going to be the same; that is what it is saying.
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Or I could pick any point, maybe here; again, from this point to point A, and this point to point B, this is going to be the same.
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That is what this one is saying.
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And then, the next one: Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment.
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This is just the converse of this; it is saying that if, without knowing where my perpendicular bisector is,
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I draw two segments from point A and from point B out to the same point,
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so that the two distances are the same, then it will lie on the perpendicular bisector.
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The first one is saying that, if the point lies on the perpendicular bisector, then it is equidistant from the two endpoints, point A and point B.
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The second one is saying that, if I draw two segments to a point equidistant from point A and point B, then it lies on the perpendicular bisector.
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OK, so then, that was the first one; that was the perpendicular bisector; the second one is a median.
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Now, when you think of median, think of middle or midpoint.
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It is kind of like a perpendicular bisector; with perpendicular bisectors, we worked with the midpoint, too.
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But that had two conditions; it had to be perpendicular to the side, and the bisector had to go through the midpoint of the side.
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Median is just through the midpoint--just that one condition.
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So, think of median as "middle"; but the condition here is this--it is going to go to the middle of the side, but from the vertex opposite that side.
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So, the vertex opposite this side is B; it can't be A, and it can't be C; it has to be B.
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So, when I draw a segment from the vertex to that point, this would be the median.
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If I label this as D, BD is the median of this triangle, of this side, AC.
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Again, the difference between this and a perpendicular bisector: a perpendicular bisector has nothing about the vertex.
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You don't care about where the vertex is.
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You just have to draw the segment so that it is perpendicular to the side, and it is cutting at the midpoint.
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The median is just the segment from the vertex to the midpoint of the side opposite.
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So then, again, since we have three sides of a triangle, I can draw three medians in a triangle.
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This is just one median; the next median I can draw from...let's say that is the midpoint--I'll do that.
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So again, I am going to draw it from this point, the segment with two endpoints; one is there, and one is at this vertex.
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And then again, from this one, let's say it is about here: 1, 2, 3, 1, 2, 3...to show that these parts are the same, or congruent.
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And then again, I am going to draw from there all the way to the vertex opposite.
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And again, for a median, if you draw all three medians of a triangle, then it should meet at one point, right there.
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That is the median; the median is like the middle; so far, we did "perpendicular bisector," and we did "median."
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OK, let's do this problem for median: Find the point S on segment AB so that CS is a median.
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I want to find a point labeled S on AB, this side right here, so that from C to that S is going to be a median.
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Now, remember: median has to do with the midpoint or middle.
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Remember: the median has two endpoints: one is from the vertex, so that means it is going to come from C, vertex C;
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and it is going to go to the midpoint of AB, and that point is going to be labeled S.
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So, I need to find the midpoint of AB, because CS has to be a median; that means S has to be the midpoint of AB.
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How do I find the midpoint when I am given two points?
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A is at 1, 2, 3...(-3,2); B is (1,-4); so I am just going 1, 2, 3, 4; positive 1, negative 4.
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To find the midpoint of this point to this point, it is going to be (x₁ + x₂) divided by 2;
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so I am going to add up the x's and divide it by 2; and then I am going to add up my y-coordinates and divide it by 2.
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So, it is like the average; to find the midpoint, you are going to find the average of the x's and the average of the y's.
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So then, for the x's, it is (-3 + 1)/2; for the y's, it is (2 + -4)/2; so this would be -2/2 and -2/2.
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Well, -2/2 is -1, and -2/2 here is -1; that means (-1,-1).
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This point right here is where S is; that means that, if I draw a line from C all the way to S, like that, that is a median,
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because S is the midpoint between A and B; so here is point S.
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The third segment is the altitude: now, altitude--just think of it as being perpendicular to the side.
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It is kind of like the perpendicular bisector, except that there is no bisector--just the "perpendicular."
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But then, one of the endpoints also has to be at the vertex.
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One point is at the vertex, and one point on the side opposite, so that it is perpendicular.
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A median only takes the "midpoint" side, and the altitude only takes the "perpendicular" side.
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And then, both of them together is like the perpendicular bisector.
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So, I just have to draw a segment going from C down to this side so that it is perpendicular, not caring about midpoint.
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All I care about is that it is from this endpoint at the vertex, and it is perpendicular to the side.
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Let's say...that right there; now again, there are three sides, so I need to draw three altitudes.
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Now, it kind of looks like this BC is already the altitude; I'll just draw it like that.
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And again, it is from this vertex to this side, so that it is perpendicular.
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And then, where are they meeting?--right there.
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So again, an altitude is from the vertex to the opposite side so that the segment is perpendicular.
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This is the fourth one, the angle bisector; now, the angle bisector is bisecting (cutting in half),
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but it is the angle that is being cut in half, not the segment, like the perpendicular bisector.
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An **angle bisector** is a segment with one endpoint on the vertex, and it is coming out;
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but for this one, because it is the angle bisector, we don't care where it lands on the side.
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It doesn't matter where it lands; it is not going to be perpendicular; it is not going to be the midpoint...
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it could be, but that is not what it has to be.
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All it has to be: the condition is that it is coming from the vertex, and it is cutting the angle in half.
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"Angle bisector" means that the angle is being cut in half.
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So, all I care about is making sure that I draw a segment so that this angle is going to be bisected.
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OK, well, let's just say that that is cut in half.
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And then, if I draw something like that, let's say that is cutting in half the angles.
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And then, if I draw that, we can say that this angle is cut in half.
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See how this has no regard for where it is touching the side (as long as it is touching it).
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But I am not saying that this has to be perpendicular, or it has to be at the midpoint--none of that.
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The only thing is that the angle has to be bisected.
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And again, I do three because there are three sides; I have to do it to each of those angles.
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And then again, they meet at one point.
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That is the angle bisector, where the segment is bisecting the angle from the vertex.
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Any point on the bisector of an angle (let me draw a triangle again; I'll say my angle bisector is right there) is equidistant from the sides of the angle.
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If I just have a point that is any point on this angle bisector, it is equidistant to the sides, like that, or like that.
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Remember: if we want to find a distance from a point to a side, it has to be perpendicular.
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Remember: if you are standing in front of a wall, how do you find the distance--how far away you are from that wall?
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You don't measure at an angle; you have to measure directly, so that you are perpendicular.
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That distance from you to the wall has to be perpendicular to the wall.
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You can't just turn your body at an angle and find your distance that way.
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The same thing works here: if you want to find the distance between this point and this side, this is you; this is the wall.
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It has to be perpendicular; so let me draw this out again so that it looks like it is perpendicular.
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You are going to go straight out like this; it is just saying that, if this is the angle bisector,
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then any point on the angle bisector is equidistant; that means that the distance to the sides of the angle,
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which are these two sides, is going to be equidistant.
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And then, this is the converse; any point on, or in the interior of, an angle, and equidistant from the sides of the angle, lies on the bisector of the angle.
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It is the same thing; it is just saying that, if I just find the distance to a point from the sides,
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so that it is equidistant, then it is going to lie on the angle bisector.
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On the angle bisector, any point is going to be equidistant from the sides.
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OK, let's go over the four that we went over, the special segments of a triangle.
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A perpendicular bisector: remember: it had to be perpendicular, and bisecting the side; that is the perpendicular bisector.
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For a median, if this is the side of a triangle, then we don't care what this looks like, as long as this is bisected--the midpoint of the side.
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The altitude: we don't care what these sides look like, as long as it is perpendicular.
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The angle bisector: it is coming out from the angle of the triangle so that these are bisected.
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A perpendicular bisector is going to be like this, like this, and like this; you can draw arrows or not; that is a perpendicular bisector.
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For a median, it is from the vertex, so that it is congruent; from the vertex...that is the median.
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The altitude is just drawn so that it is perpendicular, so it is like that.
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And the angle bisector, the last one, is drawn so that it is bisecting the angles.
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Now, this is supposed to be bisecting the angle.
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Those are the four special segments.
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Again, the perpendicular bisector has to be perpendicular and bisect the sides.
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The median is just bisecting the side; the altitude is just perpendicular; the angle bisector is bisecting the angle.
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Let's do our examples: Draw and label a figure to illustrate each: BD is a median of triangle ABC, and D is between A and C.
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BD is the median, so I am going to draw triangle ABC; there is A, B, C; and BD--that means that it is coming out from here.
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BD is the median of triangle ABC, and D is between A and C.
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That means that, since we are dealing with BD as the median, D has to be the midpoint; there is D; there is BD.
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And all you had to do is draw it and label it.
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The next one: GH is an angle bisector of triangle EFG, and H is between E and F.
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Triangle EFG: H is between E and F; GH is an angle bisector; so H is between E and F so that GH is an angle bisector.
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Now, we don't care, as long as H is anywhere in between here; it is not going to be perpendicular;
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it could be, but that is not the rule; the rule is that the angle is bisected.
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This is like this, and then this point would be H.
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So, GH is an angle bisector of that triangle.
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Draw the altitudes for each triangle: I want to draw an altitude (remember: the altitude also has an endpoint on the vertex).
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So, if I want to draw an altitude from A to side BC, it looks like this is already the altitude.
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I can just say that this side right here would be the altitude of this side BC.
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And then, B to AC is going to be like that, and then C to AB is going to be like...that is not right...let me erase part of my triangle...
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so then, this BC would be the perpendicular bisector of AB, also.
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This is going to be the altitude of BC; and then, CB is going to be the altitude of AB.
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And then, for this one, this triangle is an obtuse triangle because angle B is greater than 90 degrees.
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So, if I want to draw the altitude from this to the side, that is pretty easy; that just goes straight down.
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But then, this one is a little bit different, because I obviously can't draw an altitude
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from point A to somewhere between B and C, so that it will be 90 degrees.
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So, what I have to do is extend this out a little bit (it is kind of hard to draw a straight line on this thing).
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Let's say that that is my straight line; that is CB extending out.
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Then, my altitude will have to go outside of this triangle, because, since it is an obtuse angle,
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I would have to draw it on the outside so that it will be 90 degrees.
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If I draw it on the inside, it is just going to be a bigger obtuse angle.
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That is the altitude for that side; and then, if I want to draw the altitude from C to this side, AB, the same thing: it is an obtuse angle.
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So, I am going to extend this out; I am going to draw from C all the way there so that it is perpendicular.
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Those would be my three altitudes; now, if you want them to all meet, which they should, in this case, they all met right here.
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For this one, you would just have to keep drawing this out, keep drawing this out,
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and keep drawing this out, and then they would eventually meet right there.
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But if you just have to draw the altitudes, then you would have to just draw that, draw that, and then draw this.
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The coordinate points of triangle RST are those three points; AB is a perpendicular bisector, so it is this, through RS.
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So now, I don't have to actually graph this out on a coordinate plane.
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But if you are a visual person, and you like to see how it looks, then you can go ahead and plot them.
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I am just going to do a little sketch of what it will look like.
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So, if I have -2 right there, let's say this is R.
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S, let's say, is -2; (2,-2) is S; and T is (5,4), so here is T.
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So, my triangle is going to look something like this.
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AB is a perpendicular bisector through RS; that means that my perpendicular bisector is through this side.
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That means that this side is going to be the one that is perpendicular to it and that is bisected.
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So, that is the perpendicular bisector, which means that it is going to look something like that; it is perpendicular.
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And then, this is going to be A and B, this point and this point.
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Find the point of intersection of AB and RS.
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I want to find the point of intersection; now, first of all, we have to find the point of intersection between AB and RS.
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And then, I want to find the slope of AB and RS.
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So, the point of intersection, we know, is right there; we also know that the same point, that point A, is the midpoint of RS.
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So, as long as I find the midpoint of RS, that would be the point of intersection between the two segments,
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because again: this segment and this segment meet at point A, which is the midpoint of RS.
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The point of intersection is going to be the midpoint.
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#1: the midpoint of RS--I am going to use these two points; remember, to find the midpoint, I am going to add up my x's,
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divide it by 2, and add up my y's...6 - -2, divided by 2; it is going to be 0, comma...minus a negative becomes a plus,
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so it is plus 2, is 8; 8 divided by 2 is 4; so then, #1: the point of intersection would be (0,4).
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And again, the reason why I did midpoint is because that is where they intersect: they intersect at the midpoint of RS.
00:31:39.700 --> 00:31:43.400
And then, that is my answer; that is the point of intersection.
00:31:43.400 --> 00:31:47.800
Then, #2: Find the slope of AB and RS.
00:31:47.800 --> 00:32:04.200
I have points R and S; so to find the slope, it is (y₂ - y₁)/(x₂ - x₁); this is slope.
00:32:04.200 --> 00:32:12.900
Again, using the same points, let's see: if I label this (x₁,y₁),
00:32:12.900 --> 00:32:21.700
and label this as (x₂,y₂)...again, the x₂ and y₂
00:32:21.700 --> 00:32:38.300
is not talking about "squared"; make sure you write this 1 and these 2's below, not above it like an exponent.
00:32:38.300 --> 00:32:46.800
And this is just saying the first and the first y; the second x and second y, because we know that this point is (x,y), and this is also (x,y).
00:32:46.800 --> 00:32:49.800
So, these are the first (x,y)'s, and these are the second (x,y)'s.
00:32:49.800 --> 00:33:04.200
So, y₂ - y₁ is -2 - 6 (and this is the slope of RS);
00:33:04.200 --> 00:33:18.400
and then, x₂ is 2, minus -2, which is -8 over...this becomes plus, so it is 4; this is -2.
00:33:18.400 --> 00:33:29.300
The slope of RS is -2; now, I have to also find the slope of AB.
00:33:29.300 --> 00:33:39.800
I don't know the point for B, but I don't have to know, because, if you have two lines,
00:33:39.800 --> 00:33:45.800
and they are perpendicular to each other, then, remember: their slopes are negative reciprocals of each other.
00:33:45.800 --> 00:33:53.400
So, now that I know the slope of RS, to find the slope of AB, it is just the negative reciprocal of it.
00:33:53.400 --> 00:34:12.900
So, this is the slope of RS, and then the slope of AB is going to be the negative reciprocal; so that is the negative of (-1/2).
00:34:12.900 --> 00:34:25.400
So then, this is going to be positive 1/2.
00:34:25.400 --> 00:34:34.200
One is -2, and the other one is positive 1/2.
00:34:34.200 --> 00:34:38.700
OK, the fourth example: Draw and label the figure for the statement; then write a proof.
00:34:38.700 --> 00:34:43.800
The median to the base of an isosceles triangle bisects the vertex angle.
00:34:43.800 --> 00:34:56.700
I am going to draw and label a figure, so I need an isosceles triangle.
00:34:56.700 --> 00:35:15.800
Let's say that this is my isosceles triangle; the median to the base of it...we know that these are congruent, because it is the median...
00:35:15.800 --> 00:35:23.400
of an isosceles triangle...bisects the vertex angle; that means that we want to prove...
00:35:23.400 --> 00:35:29.800
Let me just label this; now, since this is how you want to draw and label it, you can just draw and label it however you want.
00:35:29.800 --> 00:35:38.300
So, if I label this ABC, I can label this point as D.
00:35:38.300 --> 00:35:47.500
So, my given statement--what do I know?--what is given?
00:35:47.500 --> 00:36:01.800
I have an isosceles triangle, so I can say that triangle ABC is isosceles.
00:36:01.800 --> 00:36:16.400
And then, I can say that BD is a median, because those are parts of the information that I have to use.
00:36:16.400 --> 00:36:37.300
Now, I want to prove that it bisects the vertex angle; I am going to prove that angle ABD is congruent to angle...
00:36:37.300 --> 00:36:51.600
if I said ABD, then I have to say angle CBD, because A and C are corresponding; so if I say ABD, then I have to say CBD.
00:36:51.600 --> 00:37:20.000
OK, from here, I am going to do my proof: so my statements and my reasons...#1: Triangle ABC is isosceles,
00:37:20.000 --> 00:37:30.000
and then BD is the median; and the reason for that is because it is given.
00:37:30.000 --> 00:37:39.500
2: From here, I can say that AB, this side, is congruent to this side.
00:37:39.500 --> 00:37:53.400
Now, let's see what we have to do: I am trying to prove that this angle right here is congruent to this angle right here.
00:37:53.400 --> 00:37:59.800
Now, in order for me to prove that those two angles are congruent,
00:37:59.800 --> 00:38:04.900
I would probably have to first prove that these two triangles are congruent,
00:38:04.900 --> 00:38:09.100
because there is no way that I could just say that this angle is congruent to this angle.
00:38:09.100 --> 00:38:17.400
But if I prove that these two triangles are congruent, then I can say that any two corresponding parts are congruent.
00:38:17.400 --> 00:38:22.700
So then, once these two triangles are congruent, then these two angles can be congruent.
00:38:22.700 --> 00:38:27.700
As long as they are corresponding, any two parts of the triangles are congruent.
00:38:27.700 --> 00:38:32.700
Then, I have to focus on how I am going to prove that these two triangles are congruent.
00:38:32.700 --> 00:38:39.400
Well, I know that these two sides are congruent; I know that these two sides are congruent.
00:38:39.400 --> 00:38:50.000
And I can say that this side of this triangle is congruent to this side of this triangle; that is the reflexive property.
00:38:50.000 --> 00:39:03.800
I can prove that these two triangles are congruent by SSS; if you remember the rules, there is SSS, SAS, AAS, and then Angle-Side-Angle, ASA.
00:39:03.800 --> 00:39:12.300
So, I could do that, or I have another option; I can say that, because (remember) in an isosceles triangle,
00:39:12.300 --> 00:39:20.500
if I have the two legs being congruent, then these two angles are also congruent--remember: the base angles are also congruent.
00:39:20.500 --> 00:39:25.700
So, I can say that, too, and then prove that these two triangles are congruent through SAS.
00:39:25.700 --> 00:39:31.900
Either one works; the important thing is that we prove that these two triangles are congruent,
00:39:31.900 --> 00:39:39.300
so that we can say that these two angles, those two parts, are congruent.
00:39:39.300 --> 00:39:42.200
It is up to you--do it however you want to do it.
00:39:42.200 --> 00:39:49.700
I am just going to use the reflexive property, and say that this side is congruent, and use SSS.
00:39:49.700 --> 00:40:05.500
So, I am going to say that AB is congruent to CB, and that is my side; the reason would just be "definition of isosceles triangle,"
00:40:05.500 --> 00:40:18.700
because the definition of isosceles triangle just says that two legs are congruent--"two or more sides of a triangle are congruent."
00:40:18.700 --> 00:40:22.500
And then, I am going to say that these two parts are congruent.
00:40:22.500 --> 00:40:29.900
So, even though I have it shown on my diagram, I have to write it as a step.
00:40:29.900 --> 00:40:42.900
AD is congruent to CD, and the reason for that--I am going to say "definition of median,"
00:40:42.900 --> 00:40:51.500
because it is the median that made those two parts congruent--so it is just "definition of median."
00:40:51.500 --> 00:41:03.200
And then, that is another side that I have; and then I can say, "BD is congruent to BD," and this would be BD of this triangle,
00:41:03.200 --> 00:41:08.900
and this would be BD of the other triangle; so I am saying that a side of one triangle is congruent to a side of another triangle.
00:41:08.900 --> 00:41:16.500
And that would be the reflexive property.
00:41:16.500 --> 00:41:24.600
Now, if you chose to say that angle A is congruent to angle C, then you can say "isosceles triangle theorem" as your reason--
00:41:24.600 --> 00:41:28.200
"isosceles triangle theorem" or "base angles theorem," because,
00:41:28.200 --> 00:41:33.200
since that is an isosceles triangle, automatically the base angles are congruent.
00:41:33.200 --> 00:41:39.200
And then, the fifth step...that would be another side, so then, your reason,
00:41:39.200 --> 00:41:50.500
if you say that the triangles are congruent here (the next step), wouldn't be SSS, like mine would be; it would be SAS.
00:41:50.500 --> 00:42:05.000
Triangle ABD is congruent to triangle CBD, and again, my reason is SSS.
00:42:05.000 --> 00:42:16.500
Then, from there, I can say that angle ABD is congruent to angle CBD; what is my reason?
00:42:16.500 --> 00:42:21.600
Well, see how here you proved that the triangles are congruent.
00:42:21.600 --> 00:42:38.400
Then, once this is stated, then you can say that any two parts are congruent by "corresponding parts of congruent triangles are congruent," CPCTC.
00:42:38.400 --> 00:42:41.600
So, that would be my sixth step.
00:42:41.600 --> 00:42:48.500
Again, in order to prove that these two angles are congruent, because there is no direct way to do it,
00:42:48.500 --> 00:42:53.500
I have to prove that these two triangles are congruent so that I can say
00:42:53.500 --> 00:42:56.800
that two corresponding parts, those two angles, are going to be congruent.
00:42:56.800 --> 00:43:00.700
And then, you can do that by Side-Side-Side or Side-Angle-Side.
00:43:00.700 --> 00:43:07.300
And then, once you prove that the triangles are congruent, then you can say that those two angles are congruent.
00:43:07.300 --> 00:43:13.900
When you draw and label, if it doesn't give you a figure or a diagram for it, then just draw your own.
00:43:13.900 --> 00:43:20.500
You can label it how you want, and then that would base your given and your prove statement.
00:43:20.500 --> 00:43:28.000
But as long as you write a proof for this statement that they give you, "The median to the base of an isosceles triangle
00:43:28.000 --> 00:43:32.400
bisects the vertex angle," this would be your conclusion (your "prove" statement).
00:43:32.400 --> 00:43:41.900
Your hypothesis is going to be your given, and your conclusion of your statement is your "prove" statement.
00:43:41.900 --> 00:43:44.000
Well, that is it for this lesson; I will see you next time.