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## Theorems concerning quadrilateral properties

Current time:0:00Total duration:8:30

# Proof: Opposite sides of a parallelogram

CCSS Math: HSG.CO.C.11

## Video transcript

What we're going to
prove in this video is a couple of fairly
straightforward parallelogram-related proofs. And this first one,
we're going to say, hey, if we have this
parallelogram ABCD, let's prove that the opposite
sides have the same length. So prove that AB is equal to
DC and that AD is equal to BC. So let me draw a diagonal here. And this diagonal, depending
on how you view it, is intersecting two
sets of parallel lines. So you could also consider
it to be a transversal. Actually, let me draw it a
little bit neater than that. I can do a better job. Nope. That's not any better. That is about as
good as I can do. So if we view DB,
this diagonal DB-- we can view it as a transversal
for the parallel lines AB and DC. And if you view it that
way, you can pick out that angle ABD is going to
be congruent-- so angle ABD. That's that angle
right there-- is going to be congruent
to angle BDC, because they are
alternate interior angles. You have a transversal--
parallel lines. So we know that
angle ABD is going to be congruent to angle BDC. Now, you could also
view this diagonal, DB-- you could view it as
a transversal of these two parallel lines, of the other
pair of parallel lines, AD and BC. And if you look at it that
way, then you immediately see that angle DBC
right over here is going to be
congruent to angle ADB for the exact same reason. They are alternate
interior angles of a transversal intersecting
these two parallel lines. So I could write this. This is alternate
interior angles are congruent when you have
a transversal intersecting two parallel lines. And we also see that
both of these triangles, triangle ADB and triangle CDB,
both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might
realize that we've just shown that both of
these triangles, they have this pink angle. Then they have this
side in common. And then they have
the green angle. Pink angle, side in common,
and then the green angle. So we've just shown
by angle-side-angle that these two
triangles are congruent. So let me write this down. We have shown that
triangle-- I'll go from non-labeled
to pink to green-- ADB is congruent to triangle--
non-labeled to pink to green-- CBD. And this comes out of
angle-side-angle congruency. Well, what does that do for us? Well, if two triangles
are congruent, then all of the corresponding
features of the two triangles are going to be congruent. In particular, side DC
on this bottom triangle corresponds to side BA
on that top triangle. So they need to be congruent. So we get DC is going
to be equal to BA. And that's because they
are corresponding sides of congruent triangles. So this is going to
be equal to that. And by that exact same
logic, AD corresponds to CB. AD is equal to CB. And for the exact same
reason-- corresponding sides of congruent triangles. And then we're done. We've proven that opposite
sides are congruent. Now let's go the other way. Let's say that we have some
type of a quadrilateral, and we know that the
opposite sides are congruent. Can we prove to ourselves
that this is a parallelogram? Well, it's kind of the
same proof in reverse. So let's draw a
diagonal here, since we know a lot about triangles. So let me draw. There we go. That's the hardest part. Draw it. That's pretty good. All right. So we obviously know that CB
is going to be equal to itself. So I'll draw it like that. Obviously, because
it's the same line. And then we have
something interesting. We've split this quadrilateral
into two triangles, triangle ACB and triangle DBC. And notice, all three sides
of these two triangles are equal to each other. So we know by side-side-side
that they are congruent. So we know that triangle
A-- and we're starting at A, and then I'm going
to the one-hash side. So ACB is congruent
to triangle DBC. And this is by
side-side-side congruency. Well, what does that do for us? Well, it tells us that all
of the corresponding angles are going to be congruent. So for example, angle ABC
is going to be-- so let me mark that. You can say ABC is going
to be congruent to DCB. And you could say, by
corresponding angles congruent of congruent triangles. I'm just using some shorthand
here to save some time. So ABC is going to
be congruent to DCB, so these two angles are
going to be congruent. Well, this is interesting,
because here you have a line. And it's intersecting AB and CD. And we clearly see
that these things that could be alternate interior
angles are congruent. And because we have these
congruent alternate interior angles, we know that AB
must be parallel to CD. So this must be
parallel to that. So we know that AB
is parallel to CD by alternate interior angles
of a transversal intersecting parallel lines. Now, we can use that
exact same logic. We also know that angle--
let me get this right. Angle ACB is congruent
to angle DBC. And we know that by
corresponding angles congruent of congruent triangles. So we're just saying this
angle is equal to that angle. Well, once again, these could
be alternate interior angles. They look like they could be. This is a transversal. And here's two lines
here, which we're not sure whether they're parallel. But because the alternate
interior angles are congruent, we know that they are parallel. So this is parallel to that. So we know that AC
is parallel to BD by alternate interior angles. And we're done. So what we've done
is-- it's interesting. We've shown if you
have a parallelogram, opposite sides have
the same length. And if opposite sides
have the same length, then you have a parallelogram. And so we've actually proven
it in both directions. And so we can actually make what
you call an "if and only if" statement. You could say opposite sides of
a quadrilateral are parallel if and only if their
lengths are equal. And you say if and only if. So if they are
parallel, then you could say their
lengths are equal. And only if their lengths
are equal are they parallel. We've proven it in
both directions.