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Intro to angle bisector theorem

The Angle Bisector Theorem states that when an angle in a triangle is split into two equal angles, it divides the opposite side into two parts. The ratio of these parts will be the same as the ratio of the sides next to the angle. Created by Sal Khan.

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  • female robot grace style avatar for user Jade
    What does arbitrary mean? Sal uses it when he refers to triangles and angles.
    (36 votes)
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  • male robot johnny style avatar for user Dawson
    I'm a bit confused: the bisector line segment is perpendicular to the bottom line of the triangle, the bisector line segment is equal in length to itself, and the angle that's being bisected is divided into two angles with equal measures. Based on this information, wouldn't the Angle-Side-Angle postulate tell us that any two triangles formed from an angle bisector are congruent? And yet, I know this isn't true in every case. A little help, please?
    (15 votes)
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  • purple pi teal style avatar for user Affy
    from to , I have no idea what's going on.
    (15 votes)
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  • leaf green style avatar for user Haas
    Hi, instead of going through this entire proof could you not say that line BD is perpendicular to AC, then it creates 90 degree angles in triangle BAD and CAD... with AA postulate, then, both of them are Similar and we prove corresponding sides have the same ratio.
    (7 votes)
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    • leaf green style avatar for user Matthew
      BD is not necessarily perpendicular to AC. Quoting from Age of Caffiene: "Watch out! The bisector is not [necessarily] perpendicular to the bottom line... Imagine you had an isosceles triangle and you took the angle bisector, and you'll see that the two lines are perpendicular. However, if you tilt the base, the bisector won't change so they will not be perpendicular anymore : ) "
      (12 votes)
  • blobby green style avatar for user Pragathi
    If triangle BCF is isosceles, shouldn't triangle ABC be isosceles too?
    Here's why:
    Segment CF = segment AB. CF is also equal to BC. So BC is congruent to AB. Doesn't that make triangle ABC isosceles?
    That can't be right. . . Anybody know where I went wrong?
    (6 votes)
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    • female robot grace style avatar for user Ankita Ghosh
      Unfortunately the mistake lies in the very first step....
      Sal constructs CF parallel to AB not equal to AB.
      We know that BD is the angle bisector of angle ABC which means angle ABD = angle CBD. Now, CF is parallel to AB and the transversal is BF. So we get angle ABF = angle BFC ( alternate interior angles are equal). But we already know angle ABD i.e. same as angle ABF = angle CBD which means angle BFC = angle CBD.
      Therefore triangle BCF is isosceles while triangle ABC is not.

      Hope this helps you and clears your confusion! Best wishes!! :)
      (9 votes)
  • starky sapling style avatar for user Tashu Gupta
    At , Sal says that the two triangles separated from the bisector aren't necessarily similar. This means that side AB can be longer than side BC and vice versa. My question is that for example if side AB is longer than side BC, at wouldn't CF be longer than BC? On the other hand Sal says that triangle BCF is isosceles meaning that the those sides should be the same. I understand that concept, but right now I am kind of confused.
    (6 votes)
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    • aqualine ultimate style avatar for user Timber Lin
      i think you assumed AB is equal length to FC because it they're parallel, but that's not true. imagine extending A really far from B but still the imaginary yellow line so that ABF remains constant. you can see that AB can get really long while CF and BC remain constant and equal to each other (BCF is isosceles). hope this clears things up
      (6 votes)
  • blobby green style avatar for user James Taylor
    At , what is AA Similarity? I've never heard of it or learned it before....
    (0 votes)
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  • winston baby style avatar for user Samuel Lee
    Sal mentions how there's always a line that is a parallel segment BA and creates the line. Earlier, he also extends segment BD.

    How is Sal able to create and extend lines out of nowhere? Is there a mathematical statement permitting us to create any line we want? Can someone link me to a video or website explaining my needs?
    (5 votes)
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  • blobby green style avatar for user 2027lemunyonm
    That makes no sense
    (6 votes)
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  • leafers tree style avatar for user cookjd1225
    How do I know when to use what proof for what problem? I know what each one does but I don't quite under stand in what context they are used in? this is not related to this video I'm just having a hard time with proofs in general.
    (4 votes)
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    • piceratops ultimate style avatar for user ILoveToLearn
      Hi cookid1225!
      Great question!
      I would suggest that you make sure you are thoroughly well-grounded in all of the theorems, so that you are sure that you know how to use them.
      Most of the work in proofs is seeing the triangles and other shapes and using their respective theorems to solve them. It just takes a little bit of work to see all the shapes! You might want to refer to the angle game videos earlier in the geometry course.
      For general proofs, this is what I said to someone else:
      If you can, circle what you're trying to prove, and keep referring to it as you go through with your proof. Take the givens and use the theorems, and put it all into one steady stream of logic. If you need to you can write it down in complete sentences or reason aloud, working through your proof audibly… If you understand the concept, you should be able to go through with it and use it, but if you don't understand the reasoning behind the concept, it won't make much sense when you're trying to do it.
      (5 votes)

Video transcript

What I want to do first is just show you what the angle bisector theorem is and then we'll actually prove it for ourselves. So I just have an arbitrary triangle right over here, triangle ABC. And what I'm going to do is I'm going to draw an angle bisector for this angle up here. And we could have done it with any of the three angles, but I'll just do this one. I'll make our proof a little bit easier. So I'm just going to bisect this angle, angle ABC. So let's just say that's the angle bisector of angle ABC, and so this angle right over here is equal to this angle right over here. And let me call this point down here-- let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector-- so when I put this angle bisector here, it created two smaller triangles out of that larger one. The angle bisector theorem tells us the ratios between the other sides of these two triangles that we've now created are going to be the same. So it tells us that the ratio of AB to AD is going to be equal to the ratio of BC to, you could say, CD. So the ratio of-- I'll color code it. The ratio of that, which is this, to this is going to be equal to the ratio of this, which is that, to this right over here-- to CD, which is that over here. So once you see the ratio of that to that, it's going to be the same as the ratio of that to that. So that's kind of a cool result, but you can't just accept it on faith because it's a cool result. You want to prove it to ourselves. And so you can imagine right over here, we have some ratios set up. So we're going to prove it using similar triangles. And unfortunate for us, these two triangles right here aren't necessarily similar. We know that these two angles are congruent to each other, but we don't know whether this angle is equal to that angle or that angle. We don't know. We can't make any statements like that. So in order to actually set up this type of a statement, we'll have to construct maybe another triangle that will be similar to one of these right over here. And one way to do it would be to draw another line. And this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you. What happens is if we can continue this bisector-- this angle bisector right over here, so let's just continue it. It just keeps going on and on and on. And let's also-- maybe we can construct a similar triangle to this triangle over here if we draw a line that's parallel to AB down here. So let's try to do that. So I'm just going to say, well, if C is not on AB, you could always find a point or a line that goes through C that is parallel to AB. So by definition, let's just create another line right over here. And let's call this point right over here F and let's just pick this line in such a way that FC is parallel to AB. So this is parallel to that right over there. And we could just construct it that way. And now we have some interesting things. And we did it that way so that we can make these two triangles be similar to each other. So let's see that. Let's see what happens. So before we even think about similarity, let's think about what we know about some of the angles here. We know that we have alternate interior angles-- so just think about these two parallel lines. So I could imagine AB keeps going like that. FC keeps going like that. And line BD right here is a transversal. Then whatever this angle is, this angle is going to be as well, from alternate interior angles, which we've talked a lot about when we first talked about angles with transversals and all of that. So these two angles are going to be the same. But this angle and this angle are also going to be the same, because this angle and that angle are the same. This is a bisector. Because this is a bisector, we know that angle ABD is the same as angle DBC. So whatever this angle is, that angle is. And so is this angle. And that gives us kind of an interesting result, because here we have a situation where if you look at this larger triangle BFC, we have two base angles that are the same, which means this must be an isosceles triangle. So BC must be the same as FC. So that was kind of cool. We just used the transversal and the alternate interior angles to show that these are isosceles, and that BC and FC are the same thing. And that could be useful, because we have a feeling that this triangle and this triangle are going to be similar. We haven't proven it yet. But how will that help us get something about BC up here? But we just showed that BC and FC are the same thing. So this is going to be the same thing. If we want to prove it, if we can prove that the ratio of AB to AD is the same thing as the ratio of FC to CD, we're going to be there because BC, we just showed, is equal to FC. But let's not start with the theorem. Let's actually get to the theorem. So FC is parallel to AB, [? able ?] to set up this one isosceles triangle, so these sides are congruent. Now, let's look at some of the other angles here and make ourselves feel good about it. Well, we have this. If we look at triangle ABD, so this triangle right over here, and triangle FDC, we already established that they have one set of angles that are the same. And then, and then they also both-- ABD has this angle right over here, which is a vertical angle with this one over here, so they're congruent. And we know if two triangles have two angles that are the same, actually the third one's going to be the same as well. Or you could say by the angle-angle similarity postulate, these two triangles are similar. So let me write that down. You want to make sure you get the corresponding sides right. We now know by angle-angle-- and I'm going to start at the green angle-- that triangle B-- and then the blue angle-- BDA is similar to triangle-- so then once again, let's start with the green angle, F. Then, you go to the blue angle, FDC. And here, we want to eventually get to the angle bisector theorem, so we want to look at the ratio between AB and AD. Similar triangles, either you could find the ratio between corresponding sides are going to be similar triangles, or you could find the ratio between two sides of a similar triangle and compare them to the ratio the same two corresponding sides on the other similar triangle, and they should be the same. So by similar triangles, we know that the ratio of AB-- and this, by the way, was by angle-angle similarity. Want to write that down. So now that we know they're similar, we know the ratio of AB to AD is going to be equal to-- and we could even look here for the corresponding sides. The ratio of AB, the corresponding side is going to be CF-- is going to equal CF over AD. AD is the same thing as CD-- over CD. And so we know the ratio of AB to AD is equal to CF over CD. But we just proved to ourselves, because this is an isosceles triangle, that CF is the same thing as BC right over here. And we're done. We've just proven AB over AD is equal to BC over CD. So there's two things we had to do here is one, construct this other triangle, that, assuming this was parallel, that gave us two things, that gave us another angle to show that they're similar and also allowed us to establish-- sorry, I have something stuck in my throat. Just coughed off camera. So I should go get a drink of water after this. So constructing this triangle here, we were able to both show it's similar and to construct this larger isosceles triangle to show, look, if we can find the ratio of this side to this side is the same as a ratio of this side to this side, that's analogous to showing that the ratio of this side to this side is the same as BC to CD. And we are done.