0:00:00.000,0:00:00.590


0:00:00.590,0:00:03.640
Let's say we have a circle,[br]and then we have a

0:00:03.640,0:00:05.280
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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