So given this diagram, we have to figure out what the length

of CF right over here is, and you might already guess

that this will have to do something to similar triangles

at least it looks that triangle CFE is similar to ABE

the intuition there is kind of embedded inside of it

and it also looks like triangle CFB

is going to be similar to triangle DEB

but once again we're gonna prove that to ourselves,

and then maybe we can deal with all the ratios of the different size

to CF right over here,

and actually figure out what CF is going to be

So first let's prove to ourselves

that these are definitely are similar triangles

So you have this 90 degree angle an ABE

and you know this 90 degree angle and CFE,

if we can prove just one other angle is,

or one other set of corresponding angles

is congruent in both and we've proved that they're similar

And there we can either show that,

they both show, they both share this angle over here

Angle is CEF is the same as angle AEB

So we've shown two angles, two corresponding angles

in these triangles, this is an angle on both triangles,

they are congruent, so these triangles are going to be similar

You can also show that this line, probably 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 it out of the way

We know that triangle ABE, ABE is similar with triangle CFE,

you wanna make sure get it in the right order

F is where the 90 degree angle is,

B is where the 90 degree angle is,

and so an E is where this orange angle is

So CFE, it's similar triangles CFE

Now let's see if we can figure out the same statement

going the other way, looking at triangle DEB

So once again, once again, you have 90 degree angle here

This is 90 and this is definitely gonna 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 wanna call it

So they have one set of corresponding angles that are congruent

And the you'll also see that they both share this angle right over here

On the small triangle, I'm not looking at,

I'm not looking at this triangle right over here

it's supposed to be the on the right

So they bout share this angle right over here

DBE, 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 angles, 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, so we also know,

let's scroll over to the right a little bit,

we also know that triangle, triangle DEB, triangle DEB is similar,

triangle DEB is similar to triangle CFB, to triangle CFB

Now what can we do from here?

What we know that the ratios of corresponding sites,

one of those, each of those similar triangles

are 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're only given one side

In the case of DEB and CFB we've only been given one side over here

it doesn't seem to be a lot to work with

And this is why it's slightly more a challenging problem here

let's just go ahead and see if we can assume one of the sides

and actually, maybe the sides

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 at least this give us something to work with

And Y is shared by both ABE and DEB

the shorter, the smaller triangles over there,

so maybe we'll call this length, we'll call BF X, let's call BF X

and then let's call FE, while this is X and this is Y minus X

We've introduced a bunch of,

We've introduced a bunch of variables here

but maybe with all the proportionalities of things,

just maybe things will work out,

or at least we can have a little more sense

where we can go with this actual problem

But now we can start dealing with,

we can now start dealing with the similar triangles

For example we wanna figure out what CF,

we wanna figure out what CF is

We now know that for these two triangles,

the ratio of the corresponding sides are going to be constant,

so for example the ratio between CF and 9,

the corresponding sides, the ratio between CF and 9

has got to be equal to the ratio between Y minus X, Y minus X,

that's that side right there, Y minus X

in the corresponding side of the larger triangle

While the corresponding side of the large triangle

is this entire length and that entire length over there is Y

So it's equal to Y minus X over Y

So we can simplify this a little bit,

While I'll hold off for a second,

lets see if we can do something similar with this thing on the right

not looking at triangle CFE anymore

so we have CF over DE is going to be equal to,

so CF over DE is going to be equal to X is going to be equal to,

It's going to be equal to X over, it's going to be equal to X over

this entire base right over here, this entire BE,

so which is know is over Y

And now this looks interesting because we have 3 unknowns

we have CF sorry we know what DE is already,

I could have written CF over 12

The ratio between CF and 12

is gong to be the ratio between X and Y

So we have 3 unknowns and 3 equations,

it seems it's hard to solve at first,

because its 1 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 can do a substitution so that's why this is a little tricky

So this one right here we can rewrite as CF,

let me do that same green color, we can rewrite is as CF over 9

is equal to Y minus X over Y is the same thing

as Y over Y minus X over Y or 1 minus X over Y

All I did is essentially, I could you can say distributed

then 1 over Y times both of these terms

Y over Y minus X over Y, 1 over or 1 minus X minus Y

And this is usual because we already know what Y is equal to

X, sorry, 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 homestretch here,

CF which is what we care about,

CF over 9 is equal 1 minus CF over 12

And now we have one equation and one unknown

We should be able to solve this right over here,

so we can add CF to both sides

so you have CF over 9 plus CF over 12

is equal to one, we just have to find a common denominator here

I think 36 will do the trick

So 9 times 4 is 36, so if you have to multiple 9 times 4,

you have to multiply CF times 4 so you have 4 CF,

4 CF over 36 is the same thing as CF over 9 and then plus,

CF over 12 is the same thing as over 3 CF over 36

and this is going to be equal to 1,

and then we are left with 4 CF plus 3 CF is equal to 7

CF over 36 is equal to and to solve for CF,

we can multiply both sides and the reciprocal of 7 over 36

So 36 over 7, so 36 over 7 multiply both sides times that 36 over 7

This side thing is cancelled out and we are left with,

our final, we get our drum roll now, CF is equal to, CF,

all of this stuff is cancelled out,

CF is equal to 36, 1 times 36 over 7 or just 36 over 7

And this is a pretty cool problem,

because what it shows you is you have two things

let's see this thing is some kind of a pole or a stick

or the wall of 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 12 meters tall, or 12 yards,

whatever you want, units you wanna use it,

if you wanna drape a string, 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 said how far apart

regardless of how a part they are,

the place where those two strings

will intersect are going to 36 7 tie,

I guess 5 7 tie regardless of how far they are

So that's a pretty, I don't know,

I think that was a pretty cool problem