WEBVTT

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What we're going to
prove in this video

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is a couple of fairly
straightforward

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parallelogram-related proofs.

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And this first one,
we're going to say, hey,

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if we have this
parallelogram ABCD,

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let's prove that the opposite
sides have the same length.

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So prove that AB is equal to
DC and that AD is equal to BC.

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So let me draw a diagonal here.

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And this diagonal, depending
on how you view it,

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is intersecting two
sets of parallel lines.

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So you could also consider
it to be a transversal.

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Actually, let me draw it a
little bit neater than that.

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I can do a better job.

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Nope.

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That's not any better.

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That is about as
good as I can do.

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So if we view DB,
this diagonal DB--

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we can view it as a transversal
for the parallel lines AB

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and DC.

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And if you view it that
way, you can pick out

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that angle ABD is going to
be congruent-- so angle ABD.

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That's that angle
right there-- is

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going to be congruent
to angle BDC,

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because they are
alternate interior angles.

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You have a transversal--
parallel lines.

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So we know that
angle ABD is going

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to be congruent to angle BDC.

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Now, you could also
view this diagonal,

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DB-- you could view it as
a transversal of these two

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parallel lines, of the other
pair of parallel lines, AD

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and BC.

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And if you look at it that
way, then you immediately

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see that angle DBC
right over here

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is going to be
congruent to angle

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ADB for the exact same reason.

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They are alternate
interior angles

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of a transversal intersecting
these two parallel lines.

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So I could write this.

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This is alternate
interior angles

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are congruent when you have
a transversal intersecting

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two parallel lines.

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And we also see that
both of these triangles,

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triangle ADB and triangle CDB,
both share this side over here.

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It's obviously equal to itself.

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Now, why is this useful?

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Well, you might
realize that we've just

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shown that both of
these triangles,

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they have this pink angle.

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Then they have this
side in common.

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And then they have
the green angle.

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Pink angle, side in common,
and then the green angle.

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So we've just shown
by angle-side-angle

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that these two
triangles are congruent.

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So let me write this down.

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We have shown that
triangle-- I'll

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go from non-labeled
to pink to green-- ADB

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is congruent to triangle--
non-labeled to pink to green--

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CBD.

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And this comes out of
angle-side-angle congruency.

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Well, what does that do for us?

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Well, if two triangles
are congruent, then

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all of the corresponding
features of the two triangles

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are going to be congruent.

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In particular, side DC
on this bottom triangle

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corresponds to side BA
on that top triangle.

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So they need to be congruent.

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So we get DC is going
to be equal to BA.

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And that's because they
are corresponding sides

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of congruent triangles.

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So this is going to
be equal to that.

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And by that exact same
logic, AD corresponds to CB.

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AD is equal to CB.

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And for the exact same
reason-- corresponding sides

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of congruent triangles.

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And then we're done.

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We've proven that opposite
sides are congruent.

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Now let's go the other way.

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Let's say that we have some
type of a quadrilateral,

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and we know that the
opposite sides are congruent.

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Can we prove to ourselves
that this is a parallelogram?

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Well, it's kind of the
same proof in reverse.

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So let's draw a
diagonal here, since we

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know a lot about triangles.

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So let me draw.

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There we go.

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That's the hardest part.

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Draw it.

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That's pretty good.

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All right.

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So we obviously know that CB
is going to be equal to itself.

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So I'll draw it like that.

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Obviously, because
it's the same line.

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And then we have
something interesting.

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We've split this quadrilateral
into two triangles, triangle

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ACB and triangle DBC.

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And notice, all three sides
of these two triangles

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are equal to each other.

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So we know by side-side-side
that they are congruent.

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So we know that triangle
A-- and we're starting at A,

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and then I'm going
to the one-hash side.

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So ACB is congruent
to triangle DBC.

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And this is by
side-side-side congruency.

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Well, what does that do for us?

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Well, it tells us that all
of the corresponding angles

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are going to be congruent.

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So for example, angle ABC
is going to be-- so let

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me mark that.

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You can say ABC is going
to be congruent to DCB.

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And you could say, by
corresponding angles congruent

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of congruent triangles.

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I'm just using some shorthand
here to save some time.

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So ABC is going to
be congruent to DCB,

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so these two angles are
going to be congruent.

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Well, this is interesting,
because here you have a line.

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And it's intersecting AB and CD.

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And we clearly see
that these things that

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could be alternate interior
angles are congruent.

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And because we have these
congruent alternate interior

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angles, we know that AB
must be parallel to CD.

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So this must be
parallel to that.

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So we know that AB
is parallel to CD

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by alternate interior angles
of a transversal intersecting

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parallel lines.

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Now, we can use that
exact same logic.

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We also know that angle--
let me get this right.

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Angle ACB is congruent
to angle DBC.

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And we know that by
corresponding angles congruent

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of congruent triangles.

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So we're just saying this
angle is equal to that angle.

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Well, once again, these could
be alternate interior angles.

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They look like they could be.

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This is a transversal.

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And here's two lines
here, which we're not sure

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whether they're parallel.

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But because the alternate
interior angles are congruent,

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we know that they are parallel.

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So this is parallel to that.

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So we know that AC
is parallel to BD

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by alternate interior angles.

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And we're done.

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So what we've done
is-- it's interesting.

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We've shown if you
have a parallelogram,

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opposite sides have
the same length.

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And if opposite sides
have the same length,

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then you have a parallelogram.

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And so we've actually proven
it in both directions.

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And so we can actually make what
you call an "if and only if"

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statement.

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You could say opposite sides of
a quadrilateral are parallel if

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and only if their
lengths are equal.

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And you say if and only if.

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So if they are
parallel, then you

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could say their
lengths are equal.

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And only if their lengths
are equal are they parallel.

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We've proven it in
both directions.