Let's say we have a circle, and then we have a
diameter of the circle.
Let me draw my best diameter.
That's pretty good.
This right here is the diameter of the circle or it's a
diameter of the circle.
That's a diameter.
Let's say I have a triangle where the diameter is one side
of the triangle, and the angle opposite that side, it's
vertex, sits some place on the circumference.
So, let's say, the angle or the angle opposite of this diameter
sits on that circumference.
So the triangle looks like this.
The triangle looks like that.
What I'm going to show you in this video is that
this triangle is going to be a right triangle.
The 90 degree side is going to be the side that is
opposite this diameter.
I don't want to label it just yet because that would
ruin the fun of the proof.
Now let's see what we can do to show this.
Well, we have in our tool kit the notion of an inscribed
angle, it's relation to a central angle that
subtends the same arc.
So let's look at that.
So let's say that this is an inscribed angle right here.
Let's call this theta.
Now let's say that that's the center of
my circle right there.
Then this angle right here would be a central angle.
Let me draw another triangle right here, another
line right there.
This is a central angle right here.
This is a radius.
This is the same radius -- actually this
distance is the same.
But we've learned several videos ago that look, this
angle, this inscribed angle, it subtends this arc up here.
The central angle that subtends that same arc is going
to be twice this angle.
We proved that several videos ago.
So this is going to be 2theta.
It's the central angle subtending the same arc.
Now, this triangle right here, this one right here, this
is an isosceles triangle.
I could rotate it and draw it like this.
If I flipped it over it would look like that, that, and then
the green side would be down like that.
And both of these sides are of length r.
This top angle is 2theta.
So all I did is I took it and I rotated it around to
draw it for you this way.
This side is that side right there.
Since its two sides are equal, this is isosceles, so these to
base angles must be the same.
That and that must be the same, or if I were to draw it up
here, that and that must be the exact same base angle.
Now let me see, I already used theta, maybe I'll
use x for these angles.
So this has to be x, and that has to be x.
So what is x going to be equal to?
Well, x plus x plus 2theta have to equal 180 degrees.
They're all in the same triangle.
So let me write that down.
We get x plus x plus 2theta, all have to be equal to 180
degrees, or we get 2x plus 2theta is equal to 180 degrees,
or we get 2x is equal to 180 minus 2theta.
Divide both sides by 2, you get x is equal to 90 minus theta.
So x is equal to 90 minus theta.
Now let's see what else we could do with this.
Well we could look at this triangle right here.
This triangle, this side over here also has this distance
right here is also a radius of the circle.
This distance over here we've already labeled it, is
a radius of a circle.
So once again, this is also an isosceles triangle.
These two sides are equal, so these two base angles
have to be equal.
So if this is theta, this is also going to
be equal to theta.
And actually, we use that information, we use to actually
show that first result about inscribed angles and the
relation between them and central angles subtending
the same arc.
So if this is theta, that's theta because this is
an isosceles triangle.
So what is this whole angle over here?
Well it's going to be theta plus 90 minus theta.
That angle right there's going to be theta
plus 90 minus theta.
Well, the thetas cancel out.
So no matter what, as long as one side of my triangle is the
diameter, and then the angle or the vertex of the angle
opposite sits opposite of that side, sits on the
circumference, then this angle right here is going to be a
right angle, and this is going to be a right triangle.
So if I just were to draw something random like this --
if I were to just take a point right there, like that, and
draw it just like that, this is a right angle.
If I were to draw something like that and go out like
that, this is a right angle.
For any of these I could do this exact same proof.
And in fact, the way I drew it right here, I kept it very
general so it would apply to any of these triangles.