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## Geometry (all content)

### Course: Geometry (all content) > Unit 5

Lesson 2: Quadrilateral proofs & angles- Proof: Opposite sides of a parallelogram
- Proof: Diagonals of a parallelogram
- Proof: Opposite angles of a parallelogram
- Quadrilateral angles
- Proof: Rhombus diagonals are perpendicular bisectors
- Whether a special quadrilateral can exist
- Rhombus diagonals

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# Whether a special quadrilateral can exist

Proving whether a special quadrilateral can exist or not. Created by Sal Khan.

## Want to join the conversation?

- Can a quadrilateral have a one-hundred eighty degree angle in it and have the 180 degree angle be the perpendicular bisector? I think that it would make the quadrilateral that he discusses in this video possible.(9 votes)
- If a 'quadrilateral' had one of its four angles = 180 degrees, then it would be a triangle.(10 votes)

- it violates the condition of triangle, external angleAEB is sum of internal opposite angles. so it cannot be equal to angleECB....unless angle EBC becomes zero...this makes your case, lines l and m become parrell..m i right(7 votes)
- does square have the property of being a special quadrilateral?(3 votes)
- yes it does

a square is a rectangle because a rectangles has two sets of parallel sides a nd so does a square! but a rectangle can't be a square because its sides are not all equal, so in a way a square is special(3 votes)

- If you turn that quadrilateral is it a kite?(4 votes)
- At1:10an assumption is made to prove if it is true or false the statement what is it called when you do this?(2 votes)
- It is a classic proof strategy. Lets say your given this question:

Ken's dog does not have a tail.

So for the sake of proof's, lets say that this is correct (just go with it).

If Ken's pet is a dog, the it has a tail since dog's have tails. So, this statement is wrong.

This is called a indirect proof.

Hope this helps :D(5 votes)

- What defines a quadrilateral?(2 votes)
- A quadrilateral is just a polygon with four sides. That is why it is called a "quadrilateral", since "quad" means "four".(4 votes)

- At0:43, what does it mean to put the ~ over the = sign?(1 vote)
- I wouldn't call it impossible. I'd say that it would have to be a degenerate quadrilateral (i.e., a triangle or lower).(2 votes)
- At0:43, what does it mean to put the ~ over the = sign?(1 vote)
- Wouldn't this work for a square(1 vote)
- No, I don't think so. The intersection of the transversals is a 90 degree angle, and intersection between the transversal and the outer edge of the square is a 45 degree angle. But twice the intersection of the transversal and the outer point of the square does equal the angle of the perpendicular intersection of the transversals...(2 votes)

## Video transcript

An interesting question has
come from some of the engineers here at the Khan Academy while
they were working on the code. Are there any
quadrilaterals-- And I've drawn an arbitrary
quadrilateral right over here. Quadrilateral ABCD. But is there any
quadrilateral that if I were to draw the diagonals,
so let me draw the diagonals. So one diagonal is BD. And then the other diagonal,
I'll draw it right here, is AC. And the point of intersection
of these two diagonals is, let's call that point E. Is there any quadrilateral where
angle AEB is congruent to angle ECB is congruent to this
angle right over here? ECB. So I'll let you think
about that for a second. Is there any quadrilateral
where these two angles are going to be
congruent to each other? Now to think about
whether this is possible, let's assume that it is. So let's assume that we do
have a quadrilateral where this is indeed the case. Where AEB, this angle right
over here, is congruent to angle ECB. We're just going to assume
that right from the get-go. Now let's try to visualize
this whole thing a little bit differently. DB is a segment, but it's
a segment of a larger line. So we could keep extending
it off like this. And let's call that
larger line, line l. Let me draw it like this. So this right over
here is line l. DB is a subset of line l. And CB, which is a segment,
is also a subset of a line. And we could call the line, if
we were to keep extending it, let's call that line m. So I'll draw it like this. So line m. CB is a subset of line m. And then we also
have the segment AC. And AC, once again,
is a segment but we can view it as a subset
of a larger line. And we'll call that
larger line, line n. And so let me draw
a line n like this. And we see that it intersects
both line l and line m. So let me draw it like this, so
it looks something like that. That is line n. And we've assumed that angle
AEB is congruent to angle ECB. Well point E is just
where n and l intersect. So this right over
here is point E. And point C is where
lines n and m intersect. So this is point C. So this assumption
right over here, that we're assuming
from the get-go, is saying that this
angle AEB-- I'll do it in the same magenta--
AEB is congruent to this angle. ECB is congruent to that angle. These angles are these
angles right over here. Now what does that tell
us about lines m and l? Well the way we've set it up, we
have two lines, lines l and m. Line n is a transversal. And now we have two
corresponding angles are congruent. We assumed that from
the get-go that we could find two quadrilateral,
where these two corresponding angles
are congruent. But if you have two
corresponding angles congruent like this, that means that these
two lines must be parallel. So this tells us that line
l is parallel to line m. Line l is parallel to line m. Which means that they
will never intersect. But we have a
contradiction showing up. And I'll let you think about
what that is for a second. Well the contradiction
that shows up, is that if line l is
parallel to line m, then any subset
of those two lines have to be parallel
to each other. So if l is parallel
to m, then that tells us that DB,
segment DB, needs to be parallel to segment CB,
which means that they can never intersect with each other. But that's a contradiction. They do intersect
with each other. They intersect right over here. So if we assume that
there's a quadrilateral, where this is the case, you
end up with this contradiction, something that's not possible. By definition it
was a quadrilateral. This diagonal
intersects at point B, which is where this
side intersects, too. And so they have to intersect. These two cannot be parallel in
order for this to actually even be a quadrilateral. And because of that,
it is impossible for any quadrilateral
for this angle AEB to be congruent
with this angle ECB.