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# Geometry proof problem: midpoint

Sal proves that a point is the midpoint of a segment using triangle congruence. Created by Sal Khan.

Video transcript

So we have these two parallel
lines, line segment AB and line segment CD. I should say they are
parallel line segments. And then we have
these transversals that go across them. So you have this transversal
BC right over here. And you have this
transversal AD. And what this
diagram tells us is that the distance between A
and E-- this little hash mark-- says that this line segment
is the same distance as the distance between
E and D. Or another way to think about it is that
point E is at the midpoint, or is the midpoint,
of line segment AD. And what I want to think
about in this video is, is point E also the
midpoint of line segment BC? So this is the question
right over here. So is E the midpoint
of line segment BC? And you could imagine, based
on a lot of the videos we've been seeing lately,
maybe it has something to do with congruent triangles. So let's see if we can set up
some congruency relationship between the two obvious
triangles in this diagram. We have this triangle
up here on the left. And then we have this
triangle down here. This one kind of looks
like it's pointing up. This one looks like
it's pointing down. So there's a bunch of things
we know about vertical angles and angles of transversals. The most obvious one is
that we have this vertical. We know that angle AEB
is going to be congruent. Or its measure is going to be
equal to the measure of angle CED. So we know that
angle AEB is going to be congruent to angle
DEC, which really just means they have the
exact same measure. And we know that because
they are vertical angles. Now, we also know that
AB and CD are parallel. So this line right over
here, this is a transversal. And there's actually
several ways that we can do this problem. But we know that this
is a transversal. And there's a couple of ways to
think about it right over here. So let me just continue
the transversal, so we get to see all of
the different angles. You could say that this
angle right here, angle ABE-- so this is its measure
right over here-- you could say that it is the alternate
interior angle to angle ECD, to this angle right over there. And if that didn't
jump out at you, you would say that the
corresponding angle to this one right over here is this
angle right up here. If you were to continue
this line off a little bit, these are the
corresponding angles. And then this one is vertical. But either way, angle
ABE-- let me be careful. Angle ABE is going to be
congruent to angle DCE. And we could say because it's
alternate interior angles. I'll just write a
little code here. So Alt interior angles. And then we have an
interesting relationship. We have an angle congruent to an
angle, another angle congruent to an angle. And then the next side is
congruent to the next side over here. So pink, green, side. Pink, green, side. So we can employ AAS,
angle-angle-side. And it's in the right order. So now, we know
that triangle-- we have to make sure that we get
the letters right here, that we have the right
corresponding vertices. We can say that
triangle AEB-- actually, let me start with the angle
just to make it interesting. Angle BEA, so we're starting
with the magenta angle, going to the green angle,
and then going to the one that we haven't labeled. So angle BEA, we can say,
is congruent to angle-- we start with the
magenta vertices-- C, go to the center, E, and then
go the unlabeled one, D. And we know this because
of angle-angle-side. And they correspond to each
other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know
that they are congruent, then that means corresponding
sides are congruent. So then we know these two
triangles are congruent. So that means that their
corresponding sides are congruent. So then we know that length
of BE is going to be equal-- and that's the segment
that's between the magenta and the green angles. The corresponding side is
side CE between the magenta and the green angles--
is equal to CE. And this just comes out
of the previous statement. If we number them, that's
1, that's 2, and that's 3. And so that comes
out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the
fact that BE is equal to CE. So I can mark this
off with hash. This line segment
right over here is congruent to this line
segment right over here, because we know that those
two triangles are congruent. And I've inadvertently,
right here, done a little two-column proof. This over here on the
left-hand side is my statement. And then on the right-hand
side, I gave my reason. And we're done.