So quadrilateral ABCD, they're telling us it is a rhombus,
and what we need to do, we need to prove
that the area of this rhombus is equal to 1/2 times AC times BD.
So we're essentially proving that the area of a rhombus
is 1/2 times the product of the lengths of its diagonals.
So let's see what we can do over here.
So there's a bunch of things we know about rhombi
and all rhombi are parallelograms,
so there's tons of things that we know about parallelograms.
First of all, if it's a rhombus, we know that all of the sides
are congruent.
So that side length is equal to that side length is
equal to that side length is equal to that side length.
Because it's a parallelogram, we know
the diagonals bisect each other.
So we know that this length-- let
me call this point over here B, let's
call this E. We know that BE is going to be equal to ED.
So that's BE, we know that's going to be equal to ED.
And we know that AE is equal to EC.
We also know, because this is a rhombus,
and we proved this in the last video,
that the diagonals, not only do they bisect each other,
but they are also perpendicular.
So we know that this is a right angle,
this is a right angle, that is a right angle,
and then this is a right angle.
So the easiest way to think about it
is if we can show that this triangle ADC is
congruent to triangle ABC, and if we can figure out
the area of one of them, we can just double it.
So the first part is pretty straightforward.
So we can see that triangle ADC is
going to be congruent to triangle ABC,
and we know that by side-side-side congruency.
This side is congruent to that side.
This side is congruent to that side, and they both share
a C right over here.
So this is by side-side-side.
And so we can say that the area-- so because of that,
we know that the area of ABCD is just
going to be equal to 2 times the area of,
we could pick either one of these.
We could say 2 times the area of ABC.
Because area of ABCD-- actually let me write it this way.
The area of ABCD is equal to the area of ADC plus the area
of ABC.
But since they're congruent, these two
are going to be the same thing, so it's just
going to be 2 times the area of ABC.
Now what is the area of ABC?
Well area of a triangle is just 1/2 base times height.
So area of ABC is just equal to 1/2 times the base
of that triangle times its height, which is equal to 1/2.
What is the length of the base?
Well the length of the base is AC.
So it's 1/2-- I'll color code it.
The base is AC.
And then what is the height?
What is the height here?
Well we know that this diagonal right over here,
that it's a perpendicular bisector.
So the height is just the distance from BE.
So it's AC times BE, that is the height.
This is an altitude.
It intersects this base at a 90-degree angle.
Or we could say BE is the same thing as 1/2 times BD.
So this is-- let me write it.
This is equal to, so it's equal to 1/2 times AC, that's
our base.
And then our height is BE, which we're saying
is the same thing as 1/2 times BD.
So that's the area of just ABC, that's
just the area of this broader triangle right up there,
or that larger triangle right up there,
that half of the rhombus.
But we decided that the area of the whole thing
is two times that.
So if we go back, if we use both this information
and this information right over here, we have the area of ABCD
is going to be equal to 2 times the area of ABC,
where the area of ABC is this thing right over here.
So 2 times the area of ABC, area of ABC
is that right over there.
So 1/2 times 1/2 is 1/4 times AC times BD.
And then you see where this is going.
2 times 1/4 fourth is 1/2 times AC times BD.
Fairly straightforward, which is a neat result.
And actually, I haven't done this in a video.
I'll do it in the next video.
There are other ways of finding the areas of parallelograms,
generally.
It's essentially base times height, but for a rhombus
we could do that because it is a parallelogram,
but we also have this other neat little result
that we proved in this video.
That if we know the lengths of the diagonals,
the area of the rhombus is 1/2 times the products
of the lengths of the diagonals, which is kind of a neat result.