Triangle similarity
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Similar Triangle Basics
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Similarity Postulates
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Similar triangles 1
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Similar triangles 2
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Similar Triangle Example Problems
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Similarity Example Problems
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Solving similar triangles 1
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Similarity example where same side plays different roles
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Solving similar triangles 2
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Challenging Similarity Problem
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Finding Area Using Similarity and Congruence
Challenging Similarity Problem Interesting similarity problem where we don't have a lot of information to work with
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- 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
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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