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This video is provided as supplementary material
for courses taught at Howard Community College and in this video
I'm going to talk about the side splitter theorem. I'll explain what the theorem is
about
and then I'll do a proof for it and then I wan t to talk about the converse to the theorem.
So the side split theorem says
that if a line parallel to one side of a triangle
intersects the other sides, then it divides those sides
into proportional segments. So here's what this means:
If I've got a triangle like triangle ABC
and I draw a line parallel to side BC,
we'll call that line DE, so that it intersects
the other two sides, then this theorem says
that's going to divide those sides into proportional segments.
So for this specific triangle, it would mean that
line segment AD over a line segment
DB would be the same thing, would be
equal to, line segment AD over line segment
EC. In other words, the two fractions that I just made
would be proportional to each other. So here's how we can prove this.
I've got two parallel lines DE and
BC and the two sides of the triangle
that DE intersects, could be thought of as transversals that are
cut by those two parallel lines. So we know that
corresponding angles along those transversals
would be congruent angles. So angle B
would have to be congruent with its corresponding angle ADE.
And in the same way angle C
would be congruent with its corresponding angle,
AED. The next thing I'm going to do
is draw a line from point D parallel
to side AC so that it
intersects side BC. I'll say that it
intersects it at point F.
So now I've got another pair of parallel lines,
DF and AC. And I can think of side
BC as a transversal which is cut by those parallel lines.
So that means that once again corresponding angles along that transversal
would be congruent,
so angle C would be congruent with its corresponding angle, DFB.
But we also know that angle C
is congruent with angle AED. That means
angle DFB must be congruent with angle AED.
Now let's look at triangle ADE
and triangle DBF.
We can see that these are similar triangles
because they've got two pairs of corresponding angles
that are congruent to each other. So we've got angle-angle similarity.
Now if these two triangles a similar to each other
then their sides are proportional. That would mean that side
AD over its corresponding side DB
would have the same proportion, will be the same fraction, as side AE
over its corresponding side, DF.
Now realize that
DF is congruent with
EC, because these are opposite sides of a parallelogram.
So I can take the proportion that I've just written
and instead of using DF, I can use
EC, since they are congruent. And now what I've got is
line segment AD over line segment DB
the equals line segment AE
over line segment EC. In other words, the ratios
are the same, and that's exactly the same thing I started out with.
So that's my proof of this theorem. To summarize the theorem one more time,
if I've got a triangle and I've drawn a line
parallel to one side of the triangle so that it
intersects the other two sides, then
it's going to divide those other two sides into proportional segments.
Now here's the converse to the theorem.
The converse is just turning the theorem around.
It says if a line divides two sides of a triangle
into proportional segments, then the lines parallel to the third side.
So once again I've got triangle ABC. I've drawn a line
DE through two of the sides,
through AB and AC,
and I know that it's divided those two sides
into proportional segments.
So line segment AD over line segment DB
equals line segment AE over line segment EC.
Since that's the case,
the converse of the side splitter theorem
says that line DE is parallel
to side BC. Okay,
so that's the side splitter theorem and its converse.
Take care, I'll see you next time.