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