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.

First! Thank You!

Second! You Rock!

what kind of electronic writing pad do you use for this training?

Forth! You Scissors! You Rock too!

Fifth！ u suck！

sixth! u ಠ_ಠ !

Sevens bith

isnt it impossible to manipulate the length of CF. If you have that the angle of FDC is defined and the length of DC is defined and then if Angle DCF is at point C and is parallel to line BD, doesnt that mean that the angle has to already be defined so the only way the line is equal to BC, if the shape is a rhombus or did i miss something here?

Fellow man, I call from the future. I comment in 2016.

khan sir you r a scholar

thank you!!!! i have more idea for my report!!!its so very helpful for me THANK'S

in fact you can use the ratio of areas of triangle ABD and CBD. This ratio equals to both AD/CD and AB/BC. In the first case they have the same height (the distance from B to AC) and in the second case they have the same height (the distance from D to AB and BC).

How CF=BC?

So chat

@Khan Academy can you please help me figure out the difference between a postulate and a definition's meaning in geometry please. I have test this week and I really need the help.

How we will prove the converse of angle bisector theorem

Speed 1.25x

30th

I still dont know how it works….

32th

I'm confused, isn't angle ADB = to angle CDB since it's a bisector?

This theorem can be used to derive u sub for R(x,sqrt(ax^2+bx+c)) integrals after completing the square

Thx man you just saved my day

4:55 wee wee =P jokes *sorry

I love ur videos

English bad improve it if it was good video can be understand well

cool maths, thanks khan

nice

hey kahn audio is not working

Wow!!!

69th

😏😏

why doesnt adding auxiliary lines change what you're trying to prove?

nvm, i found something from MathBitsNotebook:

"While it is never acceptable to change any of the original parts of a diagram, it is acceptable to add new lines or segments. "