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### Course: Geometry (Eureka Math/EngageNY) > Unit 2

Lesson 4: Topic C: Lessons 18-20: Similarity applications- Intro to angle bisector theorem
- Using the angle bisector theorem
- Solve triangles: angle bisector theorem
- Using similar & congruent triangles
- Use similar triangles
- Challenging similarity problem
- Geometry word problem: the golden ratio
- Geometry word problem: Earth & Moon radii
- Geometry word problem: a perfect pool shot

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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.

## Want to join the conversation?

- What does arbitrary mean? Sal uses it when he refers to triangles and angles.(37 votes)
- It just means something random. In this case some triangle he drew that has no particular information given about it.(81 votes)

- 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?(16 votes)
- Watch out! The bisector is
*not*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 : )(55 votes)

- from00:00to8:34, I have no idea what's going on.(16 votes)
- This video requires knowledge from previous videos/practices. Do the whole unit from the beginning before you attempt these problems so you actually understand what is going on without getting lost :) Good luck!(12 votes)

- 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.(8 votes)
- 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 : ) "(13 votes)

- 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?(7 votes)- 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!! :)(10 votes)

- At1:59, 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**4:37**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.(7 votes)- 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(7 votes)

- At7:02, what is AA Similarity? I've never heard of it or learned it before....(0 votes)
- If two angles of one triangle are congruent to two angles of a second triangle then the triangles have to be similar. Meaning all corresponding angles are congruent and the corresponding sides are proportional.(18 votes)

- Did Sal get some water? I cannot have my tutor dying of thirst!(8 votes)
- 3:04Sal 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)- Euclid originally formulated geometry in terms of five axioms, or starting assumptions. The first axiom is that if we have two points, we can join them with a straight line. The second is that if we have a line segment, we can extend it as far as we like.

https://en.wikipedia.org/wiki/Euclidean_geometry#Axioms(5 votes)

- That makes no sense(6 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.