Perpendicular Radius Bisects Chord

Geometry

Circles

Language and Notation of the Circle
Circles: Radius, Diameter and Circumference
Parts of a Circle
Area of a Circle
Circles and arcs
Areas of circles and sectors
Inscribed and Central Angles
Inscribed angles 1
Inscribed angles 2
Inscribed angles 3
Perpendicular Radius Bisects Chord
Right Triangles Inscribed in Circles (Proof)
Area of Inscribed Equilateral Triangle (some basic trig used)
Circles challenge

Perpendicular Radius Bisects Chord Simple proof using RSH postulate to show that a radius perpendicular to a chord bisects it

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Perpendicular Radius Bisects Chord

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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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