WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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The next lesson is on inequalities involving two triangles.
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Now, the last lesson that we did was on the triangle inequality theorem, which said that
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any two sides of a triangle had to be greater than the third side; but that was within just one triangle.
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We are going to go over a couple of inequality theorems that are based on comparing two triangles.
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The first one is the SAS inequality theorem; now, SAS sounds familiar, right?--that should sound familiar.
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We used the SAS theorem to prove that two triangles were congruent.
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So then, we had to prove, within two triangles, that a side, an angle, and another side, were congruent.
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And then, we could prove that these two triangles were congruent, and that was the SAS theorem.
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The SAS inequality theorem is a little bit different; "inequality" means that it is not equal.
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So then, we are not proving that two triangles are congruent; you would use the SAS theorem for that.
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But the SAS inequality part of it is actually a little bit different.
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The SAS inequality theorem is actually also known as the Hinge Theorem.
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And it says, "If two sides of one triangle are congruent to two sides of another triangle, and the included angle
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(the included angle was the angle between the two sides--it can't be angle A or angle C; it has to be angle B,
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the angle between the two sides) in one triangle is greater than (not congruent to) the corresponding included angle
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of the other triangle, then that side opposite, the third side, is going to be greater than the side opposite the other triangle."
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There is a little twist to it; so for this one, the side and the other side are congruent, just like in the SAS theorem.
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But the angles, the included angles of both triangles, are not congruent; one is actually going to be greater than the other.
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In this case, let's say the measure of angle B is greater than the measure of angle E; this is the included angle of this triangle,
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and this is the included angle of this triangle; so the first condition is that the two sides have to be congruent;
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the second condition is that the included angle is either less than or greater than the other included angle.
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Then, the side opposite...if this, then...AC is greater than DF.
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That is the SAS inequality theorem.
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Now, if the measure of angle B was less than the measure of angle E, then the side opposite AC, the third side, would be smaller than DF.
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That is the SAS inequality theorem; now, don't get confused between the two.
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You can either use the SAS theorem to prove that those two triangles are congruent by Side-Angle-Side,
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or you can use the SAS inequality theorem if you are trying to say that one side is greater or less than the other side.
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If this is, let's say, 10, then this has to be 10; if this is, let's say, 12, then this has to be 12.
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And if this is 60 degrees, and this is 50 degrees, then we can use the SAS inequality theorem to say that AC is then going to be greater than DF.
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The second one, the SSS inequality theorem, is the same exact thing, but you are going the opposite way now.
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It is the same thing, where you have two sides being congruent to two sides; so this S and this S are still the same.
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They have to be congruent; but then, when it comes to the third side, again, it is not congruent.
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With the third side, you are now comparing the two; and of course, this is after the two sides being congruent.
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If AC is greater than DF, then the angle opposite the measure of angle B, the greater side,
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is going to be greater than the angle opposite the shorter side.
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It is almost the same as the SAS inequality theorem; but then this time, you are actually using the sides now.
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You are using the sides, and then you are making a conclusion about the included angles.
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With the SAS inequality theorem, you used the included angles to compare, and then based that on the side opposite.
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With the SSS inequality theorem, you are using the sides to compare, and then basing that same comparison on the angles.
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We are just going to go ahead and work on our example problems.
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Write an inequality comparing the segments.
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We are going to compare AB with CD.
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Now, you can see the two triangles right here: we have triangle ABD, and we have another triangle DCB.
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We are going to compare this side, AB, with CD.
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Now, for us to use the SAS or SSS inequality theorem, we have to first know that, no matter what,
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in both triangles, the two triangles that we are comparing, two sides have to be congruent to two sides of another triangle.
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And then, it is either the A or the S that we are using the inequality for.
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So, for AD, I know that this is 20; and then BC--I know that that is 20; so then, they are automatically congruent.
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And then, BD is shared between both triangles; so then, I can just do that.
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So, if you can see this: now I have two sides of a triangle being congruent to two sides of the other triangle.
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So, side congruent; side congruent; that is one of the sides; then this side is congruent to this side of the other triangle,
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so that is the other side; and then, the included angle, we are comparing.
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So, if you can see that this is a triangle, and then this is a triangle; two sides are congruent to these two sides;
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then, this angle compared to that angle...I know that I can use the SAS inequality theorem.
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Then, using that, I am going to make a conclusion based on these sides opposite, this one with that side.
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Then, the included angle of this triangle, the measure of angle D, is less than the measure of angle...
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well, maybe I should not say the measure of angle D; I need to say that if the measure of angle ADB,
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or I could say BDA, is less than the measure of angle CBD, then I can conclude that AB, the side opposite,
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is less than...and then, what is the side opposite this one?--CD.
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AB is going to be smaller than CD, and that is SAS.
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Again, side and side must be congruent--the same thing here: this side and that side must be congruent.
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And then, based on the included angle or the other side (you are going to have inequalities comparing those two),
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either one, then you can make the conclusion about the angles opposite or the sides opposite.
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Some of them used this to make a conclusion about the sides opposite.
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OK, with CD as a median, determine if the statement is true.
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If you remember a median of a triangle, remember that the endpoint is from a vertex all the way down to the side opposite,
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so that it is hitting the midpoint; think of median as "midpoint" or "middle."
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The m gives that away: midpoint or median or middle.
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Since we know that CD is a median, I know that this AB is cut into two equal parts.
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That means that AD is congruent to BD; this and this are congruent.
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And then, since I need two triangles to compare, because of SAS and SSS inequality theorems,
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I need two triangles; so then, here are my triangles; that is triangle ACD and triangle CBD.
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I have one side taken care of; that is either this one or this one.
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And then, this side right here--if they share a side, then that is automatically congruent by the reflexive property.
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So now, I have Side (this is the included angle), Side, and this is Side, included Angle, Side.
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Now, we have both sides being congruent.
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Now, for this first one, if the measure of angle 1 is less than, is smaller than the measure of angle 2, then BC,
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which is this one right here, is greater than AC.
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This can be a little bit confusing, because it is switched around.
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If the measure of angle 1 is smaller than the measure of angle 2, well, let's flip this around, then.
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If these are flipped around, then let's flip this around.
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If the measure of angle 1 is less than the measure of angle 2, so this one is smaller than this one,
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then I can say that the measure of angle 2 is bigger, or greater, than the measure of angle 1.
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This and this are the same thing; so now, I can say that the measure of angle 2 is greater than the measure of angle 1.
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Then, BC, which is the side opposite (this is BC) is greater than the side opposite angle 1 (is AC).
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So, is that true? That is true.
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So again, the measure of angle 1 is smaller; this one is smaller, and this one is bigger.
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That means that BC, the side opposite, is bigger than AC, which is smaller.
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And then, that second one: if AC (let me just use a different color)--this is #2--is greater (this is bigger now)
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than BC (this is the smaller one), then the measure of angle 1 is smaller than the measure of angle 2?
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No, because if this is bigger, then this angle has to be also bigger.
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If this is smaller, then this one has to be smaller.
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The fact that they said that AC is greater than BC, but the measure of angle 1 is smaller than the measure of angle 2, means this one is false.
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The next example: we are going to write an inequality for x.
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Let's take a moment to look at the diagram and see what we have here.
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Well, I see two triangles; that is good, because I need two triangles to use the two inequality theorems.
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And I see that this triangle, one of my triangles, is an equilateral triangle.
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And then, I see that this triangle is an isosceles triangle.
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I need to compare my x's; now, my x here has to do with this side; my x here has to do with this side.
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So, what can I conclude about these sides (again, with SAS and SSS)?
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I need two of my sides to be congruent, so what two sides do I have to work with?
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I have this one and then...I have these two, because I have to conclude about the side opposite.
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That means that I need to use this included angle.
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And then, the same thing here happens for this side, and then this side, something with that angle, and then something with this angle.
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I have to conclude about this and about this to make it based on this side.
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If this is 40, then do we have two sides being congruent to two sides of the other triangle?
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Yes, we do, because here is my one side, and here is the side of the other triangle; so that is the side.
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Then, I have this side of this triangle, with this side of the other triangle--whichever theorem I am going to use.
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Then, I need to look at my angles to see how I can use my included angles to make a conclusion about the sides opposite.
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If this is 40, do I know what this angle measure is?
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Well, if it is equilateral, then it is also equiangular; that means that each angle is 60 degrees.
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This included angle is greater than this included angle; that means that this side opposite is going to be greater than this side opposite.
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And then, I can just simplify this; so if I subtract the x, I get x + 8 is greater than 10.
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Subtract the 8; x is greater than 2; so there is my inequality for x.
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And then, for my fourth example, I am going to write a two-column proof, and let's take a look at our given:
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AB is congruent to DC, so those two are congruent; and then, AB is also parallel to DC, so I can write those symbols there.
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The measure of angle BEC is greater than the measure of angle CED, so this is the bigger, and this is the smaller, angle.
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And then, I want to prove that BC is greater than CD.
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Well, since I know that this angle is greater than this angle, the side opposite is going to be greater than the side opposite this angle.
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But before we do that, I need to know...because all I have here is this; this is all of the information that I have,
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because that is the angle; the angle is bigger than the other angle, because I am trying to compare BC, this one, with this one;
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and if this angle opposite is bigger than this angle opposite, then this side has to be greater than this side;
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but I have to first know that this side and this side have to be congruent, because I need two sides
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to be congruent to two sides of the other triangle, and then you can base
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this whole bigger and smaller thing with the angles and the sides opposite.
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I need to say, first of all, because this is the included angle...I need to know that these two sides of this triangle
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are going to be congruent to these two sides of this triangle.
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That way, it will be SAS, and this angle will be included for this triangle, and then this will be the included angle for this triangle.
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So, how do I do that? Well, I have to basically prove...
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Now, I know that I am trying to prove that these two are going to be congruent to these two.
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Well, I know that these are already congruent to each other, because of the reflexive property.
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Now, I have to prove that BE, the other leg, or the other side, is going to be congruent to this side.
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Now, there is no way to actually prove that those sides are congruent, unless I first prove that maybe these two triangles,
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this triangle and this triangle, are congruent; then I can say, "Well, then, this side is going to be congruent to this side, because of CPCTC."
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And again, why am I doing that?--because I need to prove that this side and this side are congruent.
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It is based on these two triangles; but then it is these two sides.
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This side is already congruent to each other, so I already have one side covered.
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This side is good; but now I have to show that this side, the other side of this triangle, is congruent to the other side of this triangle, which is this.
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So, statements/reasons: my first step is to say that AB is congruent to DC; AB is parallel to DC.
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And then, the measure of angle BC is greater than the measure of angle CD; the reason is "given."
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#2: I can say that EC is congruent to EC, because of the reflexive property.
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And then, there is my side; that is one of the sides there.
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And then, I want to say...now I am going to try to prove that the triangles are congruent, these two triangles,
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so that I could say that these two corresponding sides are congruent by CPCTC.
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Since I know that BA is congruent to CD, that is one side.
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So, here, this is one side; and this is different than this side.
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This side is for the SAS inequality theorem; but for us to show that these two triangles are congruent, I am going to use this side;
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then I am going to use, let's see...now that they are parallel, I know that alternate interior angles are congruent;
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so then, I know that this angle and this angle are congruent, so angle BAE is congruent to angle...if I use BAE, then I have to use DCE.
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And then, my reason; and if you can't really see that, these are the parallel lines; that is AB and CD.
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And then, there is the transversal; I am saying that this angle right here is congruent to this angle right here.
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And that is this angle and this angle.
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If two parallel lines are cut by a transversal, then alternate interior angles are congruent.
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So then, this is a side done and an angle done, and then I know that these angles are also congruent,
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so angle BEA is congruent to angle DEC, and the reason for that is "vertical angles are congruent."
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There is another angle; so now, I can say that my triangles are congruent; triangle ABE is congruent to triangle CDE.
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And I have to make sure that it is corresponding; ADE is corresponding to CDE.
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And what was the reason for that? Angle-Angle-Side.
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And the whole point of proving that these two triangles are congruent is just so we can have this side being congruent.
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That is the only reason why we are proving that these two triangles are congruent,
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because to prove that this S is congruent takes this and this, and the only way we can prove them congruent
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is if we just prove that these two triangles are congruent, and then we say that corresponding parts of those triangles are congruent.
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So then, now I can say that BE is congruent to DE; my reason is CPCTC.
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Remember: you can only use this once you prove that the triangles are congruent.
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Then, you can say that any corresponding part is congruent.
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If you ever have to prove that two parts--maybe two sides or two angles--are congruent,
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then you have the option of doing that--proving that the triangles are congruent, and then
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saying that those corresponding parts are congruent by reason of CPCTC, Corresponding Parts of Congruent Triangles are Congruent.
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Now, we are done with this one; we have this--we are good with this--and we are good with the sides.
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Now, remember: we said that this angle is bigger than this angle; this is the bigger angle; this is the smaller angle.
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Now that we have all of our information, now we can say that BC, then, is greater than CD.
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And what was the reason? SAS inequality theorem.
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So again, this whole step right here, #1, 3, and then 4--the whole reason why we had to use that, and step 5,
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to prove that the triangles are congruent, is so that we can say that one of the sides is congruent.
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Otherwise, if these two sides are not congruent, we can't use the SAS inequality theorem.
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Even if we know that the angle is bigger than the other angle, and so the side opposite must be bigger than the other side,
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it works only if the two sides are congruent--that is why we had to prove all of that, so that we can use CPCTC.
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So, if you are still a little lost or confused with this problem, I want you to just watch the slide again.
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All that we had to do--I know it looks like a lot of work, but a big chunk of it was just to prove this pair of corresponding sides congruent,
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because we had this by the reflexive property; we have one of the sides being congruent;
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and we didn't have the other pair of sides being congruent.
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So then, that is why we had to prove the triangles congruent.
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And then, we can say, "OK, those sides are now congruent by CPCTC."
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We had this information; we had this information; and then, all of that work was to have this third piece of information.
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And then, once you have all of that, you can say your conclusion, based on the SAS inequality theorem.
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That is it for this lesson; thank you for watching Educator.com.