Proof: Opposite sides of parallelogram congruent | Quadrilaterals | Geometry | Khan Academy (2023)

Introduction

Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-quadrilaterals-theorems/v/proof-opposite-sides-of-parallelogram-congruent

Proving that a figure is a parallelogram if and only if opposite sides are congruent

Watch the next lesson: www.khanacademy.org/math/geometry/quadrilaterals-and-polygons/quadrilaterals/v/proof-diagonals-of-a-parallelogram-bisect-each-other

Missed the previous lesson?
www.khanacademy.org/math/geometry/quadrilaterals-and-polygons/quadrilaterals/v/quadrilateral-types-exercise

Geometry on Khan Academy: We are surrounded by space. And that space contains lots of things. And these things have shapes. In geometry we are concerned with the nature of these shapes, how we define them, and what they teach us about the world at large--from math to architecture to biology to astronomy (and everything in between). Learning geometry is about more than just taking your medicine ("It's good for you!"), it's at the core of everything that exists--including you. Having said all that, some of the specific topics we'll cover include angles, intersecting lines, right triangles, perimeter, area, volume, circles, triangles, quadrilaterals, analytic geometry, and geometric constructions. Wow. That's a lot. To summarize: it's difficult to imagine any area of math that is more widely used than geometry.

About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.

For free. For everyone. Forever. #YouCanLearnAnything

Subscribe to Khan Academy’s Geometry channel:
www.youtube.com/channel/UCD3OtKxPRUFw8kzYlhJXa1Q
Subscribe to Khan Academy: www.youtube.com/subscription_center

Content

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.