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Current time:0:00Total duration:9:06

CCSS.Math:

So we have a parallelogram
right over here. And what I want to prove
is that its diagonals bisect each other. So the first thing that
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 since we know that
they're parallel-- this is a
parallelogram-- we know the alternate interior
angles must be congruent. So that angle must be
equal to that angle there. And 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 lines. Now, 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 be congruent. So angle DEC must be-- so let
me write this down-- angle DEC must be congruent to angle
BAE, for the exact same reason. Now we have something
interesting, if we look at 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 proved
this to ourselves in the previous video-- that
parallelograms-- not only are opposite sides parallel,
they are 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 going to go from the blue to the
orange to the last one-- triangle ABE is congruent to
triangle-- blue, orange, then the last one-- CDE, by
angle-side-angle congruency. Now, what does that do for us? Well, we know if two
triangles are congruent, all of their
corresponding features, especially all of their
corresponding sides, are congruent. So we know that side EC
corresponds to side EA. Or I could say side AE
corresponds to side CE. They're corresponding sides
of congruent triangles, 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. Now, by the same
exact logic, we know that DE-- 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 shown 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 length. 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 are dealing with
a parallelogram. So let me see. So we're going to 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 want to prove
that this is a parallelogram. And to do that, we just
have to remind ourselves that this angle is going to
be equal to that angle-- it's one of the first things we
learned-- because they are vertical angles. So let me write this down. Let me label this point. Angle CED is going
to be equal to-- or is congruent to-- angle BEA. Well, that shows us
that these two triangles are congruent because we have
our corresponding sides that are congruent, an angle in
between, and then another side. So we now know that
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 are congruent, we know that all of the
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
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. Actually, I'll just
draw one arrow. AB is parallel to CD by
alternate interior angles congruent of parallel lines. I'm just writing
in some shorthand. Forgive the cryptic
nature of it. I'm saying it out. And so we can then
do the exact same-- we've just shown that these
two sides are parallel. We could then do
the exact same logic to show that these two
sides are parallel. And I won't necessarily
write it all out, but it's the 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. Actually, let me write it out. So we know that angle AEC
is congruent to angle DEB. They are vertical angles. And that was our reason
up here, as well. And then we see the
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 SAS congruency. Then we know that corresponding
angles must be congruent. So for example, angle CAE must
be congruent to angle BDE. And this is they're
corresponding angles of congruent triangles. So CAE-- let me do
this in a new color-- must be congruent to BDE. And now we have a transversal. The alternate interior
angles are congruent. So the two lines that the
transversal is intersecting must be parallel. So this must be
parallel to that. So then we have AC
must be parallel to be 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.