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## Geometry (all content)

### Course: Geometry (all content) > Unit 11

Lesson 5: Theorems concerning quadrilateral properties# Proof: Opposite sides of a parallelogram

CCSS.Math:

Sal proves that a figure is a parallelogram if and only if opposite sides are congruent. Created by Sal Khan.

## Want to join the conversation?

- For the second problem, couldn't you just have used the alt. int. angles theorem converse?(74 votes)
- I think so. As long as it makes sense and you get the desired conclusion, your proof is correct(26 votes)

- I still dont quite grasp the side angle thing (SSS AAA SAS ASA ASS SSA) I watched the vids explainin them a lot and still dont get it and I c thatthey keep poppin up on the other geometry vids!!! HELP?!!!?!(21 votes)
- They are just ways to identify similar or congruent triangles. The S stands for corresponding sides of equal measure on each triangle, and the A stands for corresponding angles of equal measure on each triangle. So for SSS, all three sides of a triangle would have the same lengths of all of the sides of another triangle. For AAA (or just AA, because you only need two of the angles) it would be the same thing, all three angles of a triangle would be the same as the angles on another triangle. However, because no sides must be related in this case, you are only getting similar triangles, not congruent ones. For ASA and SAS, two angles (ASA) or two sides (SAS) and the angle (for SAS) or a side (for ASA) that is surrounded by the two sides/angles; if these measures are equal to measures in the same position of another triangle, then they are congruent (an example of ASA would be at2:30). ASS and SSA don't actually work, but AAS and SAA work. For those comparisons, if two angles and a side that is not between them have the same measure as another triangle's two angles and an outside side, then both of those are congruent. It is really difficult to explain it without having any visuals, but I would have thought that the KA videos would have explained it well enough, but I haven't seen them, so I don't know.(64 votes)

- Isn't the second theorem just a converse of the first theorem ?(9 votes)
- It is just a converse of the first theorem.(7 votes)

- When labeling that the triangles are congruent by the ASA theorem. How do you identify the corresponding parts and then label?(7 votes)
- Pretend a triangle is ABC and XYZ. If you right that it means angle A = angle X, B = Y, C = Z, line AB = line XY, BC = YZ, and AC = XZ.(4 votes)

- how can we say that angle abd=bdc in the first instance?(4 votes)
- We can say that abd=bdc because the line he drew through the parallelogram is technically a transversal. Because it is a transversal, the two angles it forms are congruent, since we already know that the lines are parallel to each other. Hope this helps :)(6 votes)

- Alternate interior angles at1:02:: What are they?(3 votes)
- 'When two lines are crossed by another line (which is called the Transversal), the pairs of angles

• on opposite sides of the transversal

• but inside the two lines

are called Alternate Interior Angles.' Reference: www.mathsisfun.com(5 votes)

- What does the equal sign with the ~ mark on top mean?(3 votes)
- It is the congruent sign. It means that two shapes are congruent.(5 votes)

- I don't understand what this guy is teaching me may someone please help me understand?(3 votes)
- My suggestion to you is to watch the video, do some practice questions, and then if you don't understand try searching it up on YouTube. So you can get other people's points of view.(2 votes)

- Could you give a easy-to-understand explanation of the 'if and only if' logic used in math? (like in the example at8:16)

Any help would really be appreciated!(2 votes)- Lets talk about a square.

If it is a square, then it is a quadrilateral with all right angles and congruent sides.

If a quadrilateral has all right angles and congruent sides, then it is a square.

So both the original statement and its converse (switching the hypothesis and conclusion) are both true. Thus, we can combine it into an if and only if statement, It is a square if and only if it is a quadrilateral with all right angles and congruent sides.

It does not work if the converse is not true such as if you live in Houston, then you live in Texas. The converse is If you live in Texas, then you live in Houston which is false, there a number of counterexamples such as Dallas, Austin, etc.(4 votes)

- For the first problem couldn't you have just drawn the diagonal and found they were congruent by the SSS postulate?(2 votes)
- The SSS postulate says 'if two traingles have all three pairs of sides congruent, then the triangles are congruent.' From the given information in the first problem, we don't know that the triangles have three congruent side pairs, so we can't use that postulate.(4 votes)

## 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.