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# Challenging similarity problem

CCSS.Math:

## Video transcript

so given this diagram we need to figure out what the length of the 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 CF e is similar to a be e and the the intuition there is it's kind of embedded inside of it we're going to prove that to ourselves and it also looks like triangle C FB is going to be similar to triangle de Bie but where once again we're going to prove that to ourselves and then maybe we can deal with all the ratios of the different size to see F right over here and then actually figure out what CF is going to be so first let's let's prove to ourselves that these definitely are similar triangles so you have this 90 degree angle in a b e and you're this 90 degree angle and c fe if we can prove just one other angle is or one other set of corresponding angles is congruent in both then we've proved that they're similar and then we can either show that look they both show they both share this angle right over here angle c EF is the same as angle a eb so we've shown two angles 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 could also show that this line is proud of 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 a B II a/b is similar to triangle C Fe and you want to make sure you get in the right order F is where the 90 degree angle is B is where the 90 degree angle is so and then e is where this orange angle is so C Fe it's similar to triangle C F II now let's see if we can figure out that same statement going the other way looking at triangle de Bie so once again 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 c FB you have a 90 degree angle at D ef4 de Bie 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 not looking at I'm not looking at this triangle right over here as opposed to the one on the right so they both share this angle right over here DB e angle DB is the same as angle C bf so I've shown you already that we have this angle is congruent to this angle and we have this angle as a part of both so it's obviously course it's 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 we scroll over to the right a little bit we also know that triangle triangle de vie triangle de Bie is similar triangle de Bie is similar to triangle C FB to triangle C C F B now what can we do from here well we know that the ratios of corresponding sides for any 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 triangle so in the case of a B and C Fe we're only even given one side in the case of de b and c FB we've only been given one side right over here so there doesn't seem to be a lot to work with and 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 beside this part 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 B E 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 a B e and de B so that seems useful and then we're going to have to think about the shorter the short of the smaller the smaller triangles right over there so maybe we'll call this length we'll call B F we'll call B FX let's call VFX and then let's call Fe well if this is X and this is y minus X so we've introduced a bunch of we've introduced a bunch of variables here but maybe with all the proportionalities and things maybe 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 we can now start dealing with the similar triangles for example so we want to figure out what CF 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 nine their corresponding sides the ratio between CF and nine has got to be equal to the ratio between Y minus X Y minus X that's that side right there Y minus X and the corresponding side of the larger triangle well the corresponding side of the larger triangle is this entire 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 de Bie so when our looking at the triangle C FB not looking at triangle CF e anymore so now when you're looking at this triangle CF corresponds to de so we have CF 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 let me do that in a different color it's going to be equal to I'm using all my colors it's going to be equal to x over it's going to be equal to x over this entire base right over here so this entire B which once again we know is y so over Y and now this looks interesting because it looks like we have three unknowns we have CF sorry we know what de is already we this is 12 I could have written CF over 12 the ratio between CF and 12 is going to be the ratio between x why 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 we can rewrite as CF let me do that same green color we can rewrite it as CF over 9 is equal to Y minus x over Y is the same thing as Y over Y minus x over Y over or 1 minus X over Y all I did is I said 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 over or 1 minus X minus y and this is useful because we already know what X minus 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 that we get this is the homestretch 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 4 CF 4 CF over 36 the same thing as CF over 9 and then plus CF over 12 is the same thing as 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 7 CF 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 or seven decide things cancel out and we are left with our final we get our drumroll now CF is equal to CF so all of this stuff cancels out CF is equal to 36 1 times 36 over 7 or is just 36 over 7 and this was a pretty cool problem because what it shows you is if you have two things let's see 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 nine yards tall or nine meters tall and this over here this other one is 12 meters tall or 12 yards or whatever you want units you want to use it 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 said they're Y apart regardless of how far apart they are the place where those two strings would intersect is going to be 36 7 s high or I guess 5 and 1/7 I regardless of how far they are so that's a pretty I don't know I think that was a pretty cool cool problem