Perpendicular radius bisects chord Simple proof using RSH postulate to show that a radius perpendicular to a chord bisects it
Perpendicular radius bisects chord
- In a previous video we have already shown that if we have some circles
- centered oh right over here
- and if "OD" is a raidus, and it's a radius that bisects chord "AC"
- So bisects means that it kind of splits it into two
- Then "AB" = "BC"
- We've proved in a previous video that "OD" will be perpendicular to "AC"
- So we have proven that we can assume that it's perpendicular
- And the video that you might want to view is "For radius is perpendicular to chord
- You will hopefully find the proof of that.
- What I want to do in this video is go the other way.
- If we know that "OD" is a radius that is perpendicular to chord "AC"
- what I want to do with this video is proe that it's bisecting
- So we are not assuming that it's bisecting in this video over here.
- We are just assuming that it's perpendicular
- So essentially we are going to go the other way.
- So here we start with the fact that it bisected and we established that they were perpendicular
- That was in the previous video. Now we are going to start with the assumption that they are perpendicular
- and then prove that they bisect
- And just like what we did in the previous proof we will set up some triangles here
- since we know a lot about triangles now.
- And we will set up the triangles by drawing two more radii
- Radius "OC" and radius "OA"
- and that's useful for us because we know that they are both radii for the same circle
- So they have the same length as the radii don't change in a circle
- and you may already see where this is going because triangle "O"
- and let me label this point here, let me call it "M"
- because we are hoping that it ends up being the midpoint
- of "AC". Triangle "AMO" is a right triangle
- This is its hypotenuese. "AO" is it's hypoteneuse
- triangle "OMC" is a right triangle and this is it's hypotenuese right there
- We already showed that its hypoteneuses have the same length and both of these right triangles
- share segment or side "OM"
- So "OM" is clearly equal to itself
- and in a previous video, not the same video where we explained this thing
- In a previous video (I think) it's called "More on why SSA is not a postulate"
- We say that SSA is not a postulate, not a congruency postulate
- but we did establish in tha video that "RSH" is a congruency postulate
- and "RSH" tells us that if we have a right triangle thats where they "R" comes from
- If we have a right triangle and we have one of the sides (which ihappens to be congruent)
- and our hypotenuese is congruent. Then we have two congruent triangles
- And if you look at this right over here.
- We have two right triangles. "AMO" is a right triangle "CMO" is a right triangle
- they have one leg that is congruent right over here. "MO" and then both of their hypotenueses are congruent
- to each other. So by "RSH" we know that triangle "AMO" is congruent to triangle "CMO"
- by "RSH", by the "Rsh" postulate
- And so we know that if they are congruent then their corresponding sides have to be congruent
- So based on that we then know that "AM" is a correspinding side to "MC"
- So we know that "AM" must be equal to "MC" because they are correspinding sides
- of congruency. This implies that those are equal and if those are equal then we know that "OD" is
- Bisecting "AC" so we have established what we can do
- Anothe way that we can prove it without "RSH" is just straight up with the pythagoran theorem
- We know already from these two radii that I have set up over here that "OA"
- is equal to "OC" and we also know that "OM" is literally equal to itself
- and we also knwo from the Pythagoream Theorem that "AM" squared
- plus "OM" squared is equal to "OA" squared
- the squares of the length of the two legs squared summed up is equal to the length of the hypotenuese
- So we know that for the left triangle right over here
- "AMO" and we can set up the same relationship for "CMO"
- We know in "CMO" "CM" squared plus "OM" squared is equal to "OC" squared
- Now we know a few things
- We know that "OA" is equal to "OC"
- So for example right over here we have "OA" squared we could replace this with "OC"
- and then you can already see where this is going
- You can see that "CM" and "AM" are the same and if you want to do it a little more formally
- you can subtract "OM" squared from both sides of this equation andn you will get
- "AM" squared is equal to "OC" squared)
- (I've replaced this with "OC" squared
- Minus "OM" squared on the left hand side right over here
- and then on the right hand side
- As we subtract "OM" squared from both sides we have "CM" squared is equal to "OC" squared minus "OM"
- squared and then we can take the pricipal root of both sides of this because we really care about the
- positive root ( we do not want to have negative distances.
- So if you take the pricipal root of both sides this becomes "AM" is equal to the principle root of that
- and we also get that "CM" is equal to the principle root of that
- Well these two quantities are the same so "AM" must be equal
- to "CM"
- They are both equal to this quantity right here
- and so "CM" might be a bisector
- and this is really common sense
- If you know that two sides of two different right triangles are going to be congruent to each other you
- can always use th pythagorean theorem to get the third side and then
- the third side is uniquely constrained by the lengths of the other two sides
- because it is a right triangle
- and so these are all ways of getting to the same thing
- But now we can feel pretty good about the fact that if
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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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