Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy
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Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy

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.

About Gregory Ralls

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32 thoughts on “Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy

  1. 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?

  2. 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).

  3. @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.

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

  5. 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. "

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