Current time:0:00Total duration:9:53

0 energy points

# Challenging similarity problem

Video transcript

So given this diagram,
we need to figure out what the length of CF
right over here is. And you might already
guess that this will have to do something
with similar triangles, because at least it looks that
triangle CFE is similar to ABE. And the intuition there is it's
kind of embedded inside of it, and we're going to
prove that to ourselves. And it also looks
like triangle CFB is going to be similar
to triangle DEB, but once again, we're going
to prove that to ourselves. And then maybe we can
deal with all the ratios of the different sides
to CF right over here and then actually figure
out what CF is going to be. So first, let's prove to
ourselves that these definitely are similar triangles. So you have this
90-degree angle in ABE, and then you have this
90-degree angle in CFE. If we can prove just one
other angle or one other set of corresponding angles
is congruent in both, then we've proved
that they're similar. And there we can either show
that, look, they both share this angle right over here. Angle CEF is the
same as angle AEB. So we've shown two corresponding
angles in these triangles. This is an angle
in both triangles. They are congruent,
so the triangles are going to be similar. You can also show that this
line is parallel to this line, because obviously these
two angles are the same. And so these angles
will also be the same. So they're definitely
similar triangles, so let's just write that down. Get that out of the way. We know that triangle ABE
is similar to triangle CFE. And you want to make sure you
get it in the right order. F is where the
90-degree angle is. B is where the
90-degree angle is, and then E is where
this orange angle is. So CFE. It's similar to triangle CFE. Now let's see if we can figure
out that same statement going the other way, looking
at triangle DEB. So once again, you have
a 90-degree angle here. If this is 90, then
this is definitely going to be 90, as well. You have a 90-degree
angle here at CFB. You have a 90-degree
angle at DEF or DEB, however you want to call it. So they have one set
of corresponding angles that are congruent,
and then you'll also see that they both share
this angle right over here on the smaller triangle. So I'm now looking at this
triangle right over here, as opposed to the
one on the right. So they both share this
angle right over here. Angle DBE is the
same as angle CBF. So I've shown you
already that we have this angle is
congruent to this angle, and we have this angle
is a part of both. So it's obviously
congruent to itself. So we have two
corresponding angles that are congruent
to each other. So we know that this
larger triangle over here is similar to this smaller
triangle over there. So let me write this down. Scroll over to the
right a little bit. We also know that triangle DEB
is similar to triangle CFB. Now, what can we do from here? Well, we know that the ratios
of corresponding sides, for each of those
similar triangles, they're going to
have to be the same. But we only have one side
of one of the triangles. So in the case of ABE and CFE,
we've only been given one side. In the case of
DEB and CFB, we've only been given one
side right over here, so there doesn't seem to
be a lot to work with. And this is why this is a
slightly more challenging problem here. Let's just go ahead and see if
we can assume one of the sides, and actually,
maybe a side that's shared by both of
these larger triangles. And then maybe things
will work out from there. So let's just assume that
this length right over here-- let's just assume
that BE is equal to y. So let me just write this down. This whole length is
going to be equal to y, because this at least gives
us something to work with. And y is shared by both ABE
and DEB, so that seems useful. And then we're going to have
to think about the smaller triangles right over there. So maybe we'll call BF x. Let's call BF x. And then let's
FE-- well, if this is x, then this is y minus x. So we've introduced a
bunch of variables here. But maybe with all the
proportionalities and things maybe, just maybe,
things will work out. Or at least we'll have
a little bit more sense of where we can go with
this actual problem. But now we can start dealing
with the similar triangles. For example, so we want
to figure out what CF is. We now know that for these
two triangles right here, the ratio of the corresponding
sides are going to be constant. So for example, the
ratio between CF and 9, their corresponding
sides, has got to be equal to the
ratio between y minus x-- that's that
side right there-- and the corresponding side
of the larger triangle. Well, the corresponding
side of the larger triangle is this entire length. And that entire length
right over there is y. So it's equal to
y minus x over y. So we could simplify
this a little bit. Well, I'll hold
off for a second. Let's see if we can
do something similar with this thing on the right. So once again, we have CF,
its corresponding side on DEB. So now we're looking at
the triangle CFB, not looking at triangle CFE anymore. So now when we're
looking at this triangle, CF corresponds to DE. So we have CF over DE is going
to be equal to x-- let me do that in a different color. I'm using all my colors. It's going to be equal to
x over this entire base right over here, so this entire
BE, which once again, we know is y. So over y. And now this looks
interesting, because it looks like we have
three unknowns. Sorry, we know
what DE is already. This is 12. I could have written CF over 12. The ratio between
CF and 12 is going to be the ratio between x and y. So we have three unknowns
and only two equations, so it seems hard
to solve at first, because there's one
unknown, another unknown, another unknown, another
unknown, another unknown, and another unknown. But it looks like I can
write this right here, this expression, in
terms of x over y, and then we could
do a substitution. So that's why this
was a little tricky. So this one right here-- let me
do it in that same green color. We can rewrite it as CF over 9
is equal to y minus x over y. It's the same thing as
y over y minus x over y, or 1 minus x over y. All I did is, I essentially,
I guess you could say, distributed the 1 over y
times both of these terms. So y over y minus x over
y, or 1 minus x minus y. And this is useful, because
we already know what x over y is equal to. We already know that x over
y is equal to CF over 12. So this right over here, I can
replace with this, CF over 12. So then we get-- this is the
home stretch here-- CF, which is what we care about, CF over 9
is equal to 1 minus CF over 12. And now we have one
equation with one unknown, and we should be able to
solve this right over here. So we could add CF
over 12 to both sides, so you have CF over 9 plus
CF over 12 is equal to 1. We just have to find a
common denominator here, and I think 36
will do the trick. So 9 times 4 is 36, so if you
have to multiply 9 times 4, you have to multiply CF times 4. So you have 4CF. 4CF over 36 is the same
thing as CF over 9, and then plus CF over 12 is
the same thing as 3CF over 36. And this is going
to be equal to 1. And then we are left with
4CF plus 3CF is 7CF over 36 is equal to 1. And then to solve for
CF, we can multiply both sides times the
reciprocal of 7 over 36. So 36 over 7. Multiply both sides
times that, 36 over 7. This side, things cancel out. And we are left with--
we get our drum roll now. So all of this
stuff cancels out. CF is equal to 1 times 36
over 7, or just 36 over 7. And this was a
pretty cool problem, because what it shows you
is if you have two things-- let's say this thing is some
type of a pole or a stick or maybe the wall of a building,
or who knows what it is-- if this is 9 feet tall or 9
yards tall or 9 meters tall, and this over here,
this other one, is 12 meters tall or 12 yards or
whatever units you want to use, if you were to drape a string
from either of them to the base of the other, from the top
of one of them to the base of the other-- regardless of how far
apart these two things are going to be-- we just
decided they're y apart. Regardless of how
far apart they are, the place where those two
strings would intersect are going to be 36/7
high, or 5 and 1/7 high, regardless of
how far they are. So I think that was a
pretty cool problem.