Congruence postulates
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Congruent Triangles and SSS
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SSS to Show a Radius is Perpendicular to a Chord that it Bisects
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Other Triangle Congruence Postulates
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Two column proof showing segments are perpendicular
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Finding Congruent Triangles
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Congruency postulates
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More on why SSA is not a postulate
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Perpendicular Radius Bisects Chord
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Congruent Triangle Proof Example
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Congruent Triangle Example 2
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Congruent triangles 1
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Congruent triangles 2
SSS to Show a Radius is Perpendicular to a Chord that it Bisects More on the difference between a theorem and axiom. Proving a cool result using SSS
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- In the last video we learned that if we have two different triangles
- And if all of the corresponding sides of the two triangles have the same length,
- then by side side side we know that
- the two triangles are congruent. And I also touched a little bit on
- the idea of an axiom or a postulate.
- But I wanna be clear: sometimes you'll hear this referred to as
- a side side side theorem - theorem - and sometimes you'll hear it as
- a side side side postulate or axiom - postulate or axiom.
- And I think it's worth differentiating what these mean.
- A postulate or an axiom is something that you just assume, you assume from the get-go,
- while a theorem is something you prove using more basic,
- or using some postulates or axioms.
- So, and really in all of mathematics you make some core assumptions.
- You make some core assumptions. You call these -
- You call these the axioms or the postulates.
- Axioms - axioms or the postulates
- And then using those, you try to prove theorems.
- So maybe using that one I can prove some theorem over here
- And they'd be using that theorem
- and then this axiom, I can prove another theorem over here
- and them using both of those theorems
- I can prove another theorem over here - I think you get the picture.
- This axiom might lead us to this theorem
- And these two might lead us to this theorem right over here
- And we essentially try to build our knowledge
- Or we build a mathematics around these core assumptions.
- In an introductory geometry class, we kind of
- We don't rigorously prove side side side
- We don't rigorously the prove side side side theorem
- And that's why a lot of geometry calls you kind of just take it as a given
- As a postulate or an axiom
- And the whole reason why I'm doing this is
- One I just want you to know the differences between the words
- theorem and postulate or axiom
- And also so that you don't get confused. It is just a given
- but in a lot of books and I've looked at several books
- they do refer to it as a side side side theorem
- even though they never prove it rigorously.
- They do just assume it.
- So it really is more of a postulate or an axiom.
- Now with that out of they way, we just take -
- we're just going to assume going forward that we just know
- that this is true, we're gonna take it as a given
- I wanna show you that we can already do something pretty useful with it.
- So let's say that we have a circle.
- Let's say we have a circle. There's many useful things
- that we can already do with it.
- And this circle has a center right here at A
- and let's say that that we have a chord
- a chord in this circle that that is not a diameter.
- So let me draw a chord here.
- So let me draw a chord in the circle, so that's a kind of a segment
- of a secant line, and let's say that I have
- Let's say that I have a line that bisects - that bisects this chord from - from the center.
- And I guess I can call it a radius 'cause I'm gonna go from
- the center to - to the edge of the circle right over there.
- So I went to the center to circle itself.
- And when I say bisects it, these are all
- I'm just setting up the problem right now.
- When I say bisecting it, it means it splits that line segment in half.
- So what it tells us is, is that the length of this segment right over here
- is going to be equivalent to the length of this segment right over there.
- And what I wanna do is, so this is, I set it up I have a circle.
- This radius bisects this chord right over here.
- And what I wanna do is prove - the goal here is to prove, is to prove
- that it bisects this chord at a right angle.
- Or another way to say it, let me add some points here
- Let's just call this B, let's call this C, and let's call this D.
- I wanna prove that segment AB, segment AB is perpendicular
- It intersects it at a right angle, it is perpendicular to segment CD, to segment CD.
- And as you can imagine I'm gonna prove it pretty much
- using the side side side whatever you wanna call it
- side side side theorem, postulate or axiom.
- So let's do it, let's think about it this way.
- So you can imagine if I'm gonna use this I need to have some triangles,
- there's no triangles here right now, but I can construct triangles,
- and I can construct triangles based on things I know.
- For example, I can construct - this has some radius
- So let's call this, that's a radius right over here, the length of that is just going to be the radius of the circle,
- but I can also do it right over here, the length of AC
- is also going to be the radius of the circle.
- So we know that these two lines have the same length
- which is the radius of the circle, which is the radius of the circle
- or we can say that AD is congruent to AC
- or the have the exact same lengths.
- We know from the set up in the problem
- we know from the set up in the problem
- that this segment is equal in length to this segment over here.
- We could even, let me add a point here so I can refer to it
- so if I call that point E we know from the set up in the problem
- that CE is congruent to ED or they have the same length
- CE has the same length as ED.
- And we also know that both of these triangles - the one here on
- the left and the one here on the right, the one here on the right
- the both share the side EA so EA is clearly equal to EA.
- So this is clearly equal to itself, it's the same side.
- The same side has been used for both triangles.
- The triangles are adjacent to each other.
- And so we see a situation where we have a tri - where we have two different triangles
- that have corresponding sides being equal.
- This side is equivalent to this side right over here.
- This side is is equal in length to that side over there.
- And then we have obviously AE is equivalent to itself
- It's a side on both of them.
- It's a corresponding side on both of these triangles.
- And so by side side side, so by side side side
- we know, we know that triangle, triangle ABC
- triangle ABC is congruent to triangle AE -
- Oh sorry, it's not ABC it's AEC, sorry.
- We know, it should let me write it over here
- by side side side we know, we know that triangle AEC, AEC
- is congruent to triangle AED.
- But how does that help us? How does that help us knowing
- that you know we use our little theorem
- but how does that actually help us here?
- Well what's cool is once we know that two triangles are congruent
- So because, so because they congruent that tells us
- So from that we can deduce that all the angles are the same.
- And in particular we can deduce that this angle right over here
- that the measure of angle CEA, CEA is equivalent
- to measure of angle DEA, DEA, the measure of angle DEA.
- And the reason why that's also, why that's useful
- is that we also see just by looking at this
- that they are supplementary to each other.
- They're adjacent angles, their outer sides form a straight angle.
- So CEA is supplementary and equivalent to DEA.
- So they're also supplementary. So we also have the measure
- of angle CEA, the measure of angle CEA
- plus the measure of angle DEA is equal to 180 degrees.
- But they're equivalent to each other so I can replace
- the measure of DEA with the measure of CEA, measure of angle CEA.
- Or I could rewrite this as two times
- the measure of angle CEA is equal to 180 degrees.
- Or I could divide both sides by two and I'd say the measure
- of angle CEA is equal to 90 degrees
- which is going to be the same as the measure of angle DEA, because they're equivalent.
- So we know that this angle right over here is 90 degrees
- so I can glue it with that little box
- and this angle right over here is 90 degrees
- and because AB intersects where, intersects CD we have a
- 90 degree angle here and there
- and we could also prove this over there as well
- they are perpendicular to each other.
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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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