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Current time:0:00Total duration:8:19

CC Math: HSG.C.A.2

In the last video,
we learned that if we have two different
triangles and if all of the corresponding
sides of the two triangles have the same length,
then by side-side-side, we know that the two
triangles are congruent. And I also touched a
little bit on the idea of an axiom or postulate. But I want to be clear. Sometimes you will
hear this referred to as a side-side-side
theorem, and sometimes you'll hear it as a side-side-side
postulate or axiom. And I think it's
worth differentiating what these mean. A postulate or an
axiom is something that you just assume, you
assume from the get-go, while a theorem is something
you prove using more basic or using some
postulates or axioms. Really in all of mathematics,
you make some core assumptions. You call these the
axioms or the postulates. And then using those, you
try to prove theorems. So maybe using that one, I can
prove some theorem over here. And maybe using that
theorem, and then this axiom, I can prove another
theorem over here. And then using both
of those theorems, I can prove another
theorem over here. I think you get the picture. This axiom might lead
us to this theorem. These two might lead us to
this theorem right over here. And we essentially try
to build our knowledge or we build a mathematics
around these core assumptions. In an introductory
geometry class, we don't rigorously prove
the side-side-side theorem. And that's why in a lot
of geometry classes, you kind of just
take it as a given, as a postulate or an axiom. And the whole reason
why I'm doing this is one, just so you know the
difference between the words theorem and postulate or axiom. And also so that you
don't get confused. It is just a given,
but in a lot of books-- and I've looked
at several books-- they do refer to it as the
side-side-side theorem, even though they never
prove it rigorously. They do just assume it. So it really is more of
a postulate or an axiom. Now with that out
of the way, we're just going to
assume going forward that we just know
that this is true. We're going to
take it as a given. I want to show you
that we can already do something pretty
useful with it. So let's say that
we have a circle. And there's many useful things
that we can already do with it. And this circle has a
center, right here at A. And let's say that we have
a chord in the circle that is not a diameter. So let me draw a chord here. So let me draw a
chord in this circle. So it's kind of a
segment of a secant line. And let's say that
I have a line that bisects this chord
from the center. And I guess I call it
a radius because I'm going to go from the center
to the edge of the circle right over there. So I'm going to the center
to the circle itself. And when I say bisects
it-- so these are all-- I'm just setting up
the problem right now. When I say bisecting
it, it means it splits that line
segment in half. So what it tells is, is that
the length of this segment right over here is going to
be equivalent to the length of this segment
right over there. I've set it up. I have a circle. This radius bisects this
chord right over here. And the goal here
is to prove that it bisects this chord
at a right angle. Or another way to say it--
let me add some points here. Let's call this B.
Let's call this C, And let's call this
D. I want to prove that segment AB
is perpendicular. It intersects it
at a right angle. It is perpendicular
to segment CD. And as you could
imagine, I'm going to prove it pretty much using
the side-side-side whatever you want to call it, side-side-side
theorem, postulate, or axiom. So let's do it. Let's think about it this way. So you can imagine if
I'm going to use this, I need to have some triangles. There's no triangles
here right now. But I can construct triangles,
and I can construct triangles based on things I know. For example, I can construct--
this has some radius. That's a radius right over here. The length of that is just going
to be the radius of the circle. But I can also do
it right over here. The length of AC is also going
to be the radius of the circle. So we know that these two lines
have the same length, which is the radius of the circle. Or we could say that
AD is congruent to AC, or they have the
exact same lengths. We know from the
set-up in the problem that this segment
is equal in length to this segment over here. Let me add a point here
so I can refer to it. So if I call that
point E, we know from the set-up in the problem,
that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of
these triangles, the one here on the left and the one here
on the right, they both share the side EA. So EA is clearly equal to EA. So this is clearly
equal to itself. It's the same side. The same side is being
used for both triangles. The triangles are
adjacent to each other. And so we see a situation where
we have two different triangles that have corresponding
sides being equal. This side is equivalent to
this side right over here. This side is equal in length
to that side over there. And then, obviously, AE
is equivalent to itself. It's a side on both of them. It's the corresponding side
on both of these triangles. And so by side-side-side, AEC. Let me write it over here. By side-side-side, we
know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help
us knowing that we used our little theorem? But how does that
actually help us here? Well, what's cool is once we
know that two triangles are congruent-- so because they
are congruent, from that, we can deduce that all
the angles are the same. And in particular,
we can deduce that this angle right over here,
that the measure of angle CEA is equivalent to the
measure of angle DEA. And the reason why that's
useful is that we also see, just by looking
at this, that they're supplementary to each other. They're adjacent angles. Their outer sides
form a straight angle. So CEA is supplementary
and equivalent to DEA. So, they're are
also supplementary. So we also have the
measure of angle CEA plus the measure of angle
DEA is equal to 180 degrees. But they're equivalent
to each other. So I could replace the measure
of DEA with the measure of CEA. Or I could rewrite this as 2
times the measure of angle CEA is equal to 180 degrees. Or I could divide
both sides by 2. And I say the
measure of angle CEA is equal to 90
degrees, which is going to be the same as the
measure of angle DEA because they're equivalent. So we know that this angle
right over here is 90 degrees. So I can do it with
that little box. And this angle right
over here is 90 degrees. And because AB intersects
where it intersects CD, we have a 90 degree
angle here and there. And we could also prove
this over there as well. They are perpendicular
to each other.