So, we have a parallelogram right over here

So, what we wanna prove is that it's diagonals bisect each other

So, the first thing we can think about; these aren't just diagonals,

these are lines that are intersecting parallel lines

So, you can also view them as transversals

And if we focus on DB right over here, we see that it intersects DC

and AB and it sits there

Those we know are parallelograms

We know that they are parallel

This is a parallelogram

Alternate interior angles must be congruent

So, that angle must be equal to that angle there

Let me make a label here

Let me call that middle point E

So, we know that angle ABE must be congruent to angle CDE

by alternate interior angles of

a transversal intersecting parallel line

Alternate interior angles

If we look at diagonal AC or we should call it transversal AC

we can make the same argument

It intersects here and here

These two lines are parallel

So, alternate interior angles must congruent

So, angle DEC must be --- let me write this down ---

angle DEC must be congruent to angle BAE

by for the exact same reason

Now we have something interesting

If we look this top triangle over here and this bottom triangle,

we have one set of corresponding angles that are congruent

We have a side in between that's going to be congruent

Actually, let me write that down explicitly

We know and we've proved this to ourselves in the previous video

that parallelograms not only are opposite sides are parallel they're

also congruent

So, we know from the previous video that that side is equal

to that side

So, let me go back to what I was saying

We have two sets of corresponding angles that are congruent

We have a side in between that's congruent

And then we have another set of corresponding angles

that are congruent

So, we know that this triangle is congruent to that triangle

by angle-side-angle

So, we know that triangle --- I'm gonna go from the blue

to the orange to the last one

Triangle ABE is congruent to triangle blue, orange

and the last one, CDE by angle-side-angle congruency

Now what is that do for us

What we know if two triangles are congruent, all of their

corresponding features especially all of the corresponding

sides are congruent

So, we know that side EC corresponds to EA

Or I could say side AE, we could say side AE,

corresponds to side CE

They're corresponding sides of congruent triangle

So, their measures or their lengths must be the same

So, AE must be equal to CE

Let me put two slashes since I already used one slash over here

let me focus on this -- we know that BE must be equal to DE

Once again they're corresponding sides of two congruent triangles

so they must have the same length

So, this is corresponding sides of congruent triangles

So, BE is equal to DE

And we've done our proof

We've showed that, look, diagonal DB is splitting AC into two

segments of equal length and vice versa

AC is splitting DB into two segments of equal lengths

So, they are bisecting each other

Now, let's go the other way around

Let's prove to ourselves that if we have two diagonals

of a quadrilateral that are bisecting each other that we're

dealing with a parallelogram

So, let me see

So, we're gonna assume that the two diagonals

are bisecting each other

So, we're assuming that that is equal to that

And that that, right over there, is equal to that

Given that we wanna prove that this is a parallelogram

And to do that we just have to remind ourselves

We just have to remind ourselves that this angle is going

to be equal to that angle

One of the first things we learn because they're vertical angles

So, let me write this down

C -- label this point -- angle CED is going to be equal to

or is congruent to angle, so I started is BEA, angle BEA

And that, what is that, well that shows us that these

two triangles are congruent 'cause we have a corresponding sides

of a congruent and angle in between and on the other side

So, we now know that the triangle, I'll keep this in yellow,

triangle AEB is congruent to triangle DEC by side-angle-side

congruency, by SAS congruent triangles

Fair enough

Now, if we know that two triangles congruent we know that all

corresponding sides and angles are congruent

So, for example, we know that angle CDE is going to be congruent

to angle BAE

And this is just corresponding angles of congruent triangles

And now we have this kind of transversal of these two lines that

could be parallel if the alternate interior angles are congruent

And we see that they are

These two are kind of candidate alternate interior angles and

they are congruent

So, AB must be parallel to CD

So, AB, let's just draw one arrow, AB must be parallel to CD

by alternate interior angles congruent of parallel lines

I'm just writing in some short hand, forgive the cryptic nature

of it although I'm saying it out

And so we can then do the exact same -- while we just shown

that these two sides are parallel -- we can do that exact same

logic to show that these two sides are parallel

I won't necessarily write it all out

It's exact same proof to show that these two

So, first of all, we know that this angle is congruent to that angle

right over there

And then we know, actually let me write it out, we know

that angle AEC is congruent to angle DEB, I should say

They are vertical angles

And that is the reason up here as well

Vertical angles

And then we see that triangle AEC must be congruent

to triangle DEB by side-angle-side

So, then we have triangle AEC must be congruent to triangle

DEB by SAScongruency

Now, we know that corresponding angles must be congruent

So, that we know that angle, so, for example angle CAE

must be congruent to angle BDE and this is the corresponding

angles of congruent triangles

So, CAE, let me use a new color

CAE must be congruent to BDE

And now we have a transversal

The alternate interior angles are congruent

So, the two lines that the transversals are intersecting

must be parallel

So, this must be parallel to that

So, then we have AC must be parallel to BD

by alternate interior angles

And we're done

We've just proven that if the diagonals bisect each other,

if we start that as a given then we end at a point where we say,

"Hey, the opposite sides of this quadrilateral must be parallel

or that ABCD is a parallelogram "