Challenging similarity problem

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.