WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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For this lesson, we are going to go over inscribed angles of circles.
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An **inscribed angle** is an angle within a circle whose vertex is on the circle.
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We know that the sides of the inscribed angles are chords (remember: chords, again, are segments whose endpoints lie on the circle).
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So, the vertex and the sides of an inscribed angle are on the circle...and this is the angle, right there.
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Remember: if this is the angle, then this arc that it is hugging is the intercepted arc.
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From this point all the way to here--this arc is intercepted by this inscribed angle.
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The inscribed angle is half the measure of the intercepted arc.
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Now, we went over central angles; we know that central angles are angles whose vertex is on the center.
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Be careful with central angles and inscribed angles.
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The arc with the central angle are congruent; we know that the intercepted arc and the central angle have the same measure.
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But the intercepted arc with the inscribed angle is different; the arc has double the measure of the inscribed angle.
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If this arc measure, let's say (the measure of arc AB), is θ, then the measure of angle ACB is θ divided by 2.
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This angle is half the measure of this arc; the measure of angle ACB is half the measure of the arc.
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You just take the arc measure, and you divide it by 2 to get the inscribed angle.
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Be careful: the biggest mistake with this, the most common mistake, would be: if the arc, let's say, is 100,
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then this angle is 50; be careful that this is the bigger measure.
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I have seen...if this is 100, then sometimes students make that mistake and make this 200--multiply it by 2,
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thinking that this arc is half the measure of the angle; don't get that confused.
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Make sure that the intercepted arc is double the angle, or that the angle is half the arc.
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If this is θ, then the arc will be 2θ; if the arc is θ, then this would be 1/2θ.
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Always just remember that the arc is bigger than the inscribed angle.
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Now, if we have two inscribed angles with the same intercepted arc--here we have this black angle right here--
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that is the inscribed angle; the intercepted arc is from here to here; and we have another inscribed angle, the red angle,
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using the same intercepted arc; well, let's say that this has a measure of 100; if the arc has a measure of 100,
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then this inscribed angle is 50 degrees; then this inscribed angle is also 50 degrees,
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because it is 1/2 the measure of the intercepted arc, and they are both intercepting the same arc.
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So then, these have to be congruent; all inscribed angles with the same intercepted arc are congruent.
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Again, they are both inscribed angles with the same intercepted arc.
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That means that these angles have to be congruent.
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OK, if an inscribed angle of a circle intercepts a semicircle, then the angle is a right angle.
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We know that, let's say, if you draw a diameter right there, that makes this a semicircle.
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This semicircle has a measure of 180; this is the inscribed angle with this intercepted arc.
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So, the inscribed angle intercepts a semicircle; since we know that the inscribed angle is half the measure of the intercepted arc,
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well, if the intercepted arc is 180, then the inscribed angle has to have a measure of 90.
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Remember: the inscribed angle is half, so if this is 180, then this has to be 90, which makes this a right angle.
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Again, a semicircle is 180; half of that is 90, which makes that the measure of the inscribed angle; therefore, it is a right angle.
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An **inscribed quadrilateral**: now, we know that an inscribed polygon is when a polygon is inside a circle,
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with all of the vertices touching a circle; so here, we have a quadrilateral...
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Now, this is not a rectangle; we know that it is nothing special; it is just a quadrilateral, inscribed.
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If any type of quadrilateral is inscribed in a circle, opposite angles (angle B with angle D are opposite angles) are supplementary.
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And angle C with angle A are going to be supplementary.
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The measure of angle B, plus the measure of angle D, is going to equal 180.
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The measure of angle C with the measure of angle A is also going to be 180.
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We know that all four angles of a quadrilateral always add up to 360.
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So then, these two, B and D, will add up to 180; and C and A will add up to 180.
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Our examples: the first one: Name the central angle, the inscribed angle, and the intercepted arc.
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Here, there are a couple of central angles here; but they are asking for the central angle and the inscribed angle that share the same intercepted arc.
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Now, it doesn't matter; you can just name any central angle.
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Here, I know that the central angle is when the angle is inside a circle with the vertex at the center ("central"--think of "center").
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This one and this one are always confused a lot, so be careful; with central, the vertex is on the center;
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an inscribed angle is the angle where the vertex is on the circle.
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Central angle and intercepted arc have the same measure; the inscribed angle is half the measure of the intercepted arc.
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Here is the central angle; the central angle is angle CPB.
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I can also say angle APB--that is another one.
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An inscribed angle is, again, an angle whose vertex is on the circle; that would be right there, angle CAB.
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And then, the intercepted arc for each one: for the central angle CPB, the intercepted arc would be CB; that is the arc.
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The intercepted arc for this one is arc AB; and then, for this one, it is going to be arc CB.
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Find the value of x: the first one: we have a circle with an inscribed angle, and that is x; that is what we are looking for.
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This side is a semicircle; this side is, also; so this intercepted arc with this arc is 92 degrees.
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Since I know that the inscribed angle is half the measure of the intercepted arc, as long as I can find the measure of this arc, I can find x.
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I am going to take 180 (because a semicircle is 180: this is 180, but I don't have to worry about that), minus the 92; then I will get this arc.
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So, once I find this arc, then I can just divide that by 2 and get the inscribed angle.
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180 - 92 is going to be 88; if this is 88, then what is x?
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x is 88 times 2, or divided by 2? Well, remember again: the angle is smaller than the arc, so the inscribed angle is half.
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So, it would be x = 88/2, which is 44; so that right there, that inscribed angle, has a measure of 44 degrees.
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And the next one: they give us three angles of the inscribed quadrilateral in a circle.
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I know that opposite (now, be careful here; it is not consecutive--it is opposite) angles of an inscribed quadrilateral are supplementary.
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If I want to find x, then I have to use this angle and this angle here.
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Be careful that you don't do 98 + this angle; again, it is not consecutive angles that are supplementary.
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3x - 6 + x + 18 is going to be 180; so here, I have 4x + 12 is going to equal 180; 4x...if I subtract 12, I am going to get 168;
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and then, if I divide the 4, then I am going to get 42; so x is 42.
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Now, all they are asking for is the value of x, so that would be the answer.
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But if they were asking us for the actual angle measure, then we would have to plug this number back in and solve for the angle.
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This is not the angle measure; this is just x; so you would take the 42 and plug it back into this;
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and for this angle measure, you are going to get 42 + 18 (I will just do that in red); 42 + 18 is going to give us 60 degrees; this one is 60.
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And then, this one here: I can just subtract it from 180, because again, supplementary is what we used to find x.
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You can say that this is going to be 180 - 60, which is 120; or if you want to just double-check your answer,
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and just solve it out, even though you know it is going to be 120; just plug in x of 42 - 6.
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3 times 42 is 126, minus the 6 is 120; so we know that that is correct.
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For the next example, we are going to find the measure of each of these.
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Here, I have the measure of arc DC, 60 degrees; I have, let's see, chords; I have a diameter; I have a radius; I have inscribed angles.
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Here is an inscribed angle; here is an inscribed angle...central angles.
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The first one: the measure of angle CPD--now, since this is the center, this angle, we know, is called a central angle,
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because the vertex is on the center of the circle; so, central angles have the same measure as the intercepted arc.
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If the intercepted arc is DC, and it has the measure of 60, then the measure of angle CPD has to also be 60 degrees.
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The next one: the measure of arc BAC--it is this major arc here; we know that it is a major arc, because it gives us BAC, not just BC.
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If they said BC, then it would be this arc right here, BC, the minor arc; but BAC is all the way around here.
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And then, to find the measure of that...well, I have this little piece right here; do I have this and this?
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Well, I don't what the measure of arc BA is; I don't know what the measure of arc AD is; but I know that together, this whole thing
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is going to be 180, because it is a semicircle; so 180 + 60 is going to give us the measure of arc BAC;
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so this is going to be 180 + 60, which is going to be 240.
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And the last one, the measure of angle BAC: BAC is this angle; it is BAC, so this angle right here is what they are looking for.
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We know that that is an inscribed angle; if that is an inscribed angle, the intercepted arc would be arc BC;
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all of this right here is the intercepted arc for this angle here; so do we know the measure of the intercepted arc?
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Well, they didn't tell us, but we can find it, because BCD, that arc, is a semicircle.
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So, if this is 60, then this has to be 100 minus the 60; so this will be 120.
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Then, the central angle here, BPC, is going to also be 120.
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And then, what about this angle BAC--is it 2 times the intercepted arc, or half the intercepted arc?
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We know that the inscribed angle is half the arc; so if this is 120, then the measure of angle BAC has to be 120/2, which is 60 degrees.
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And the last example is going to be a proof: here is the center--the center is at P; what is given?
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Arc BC is congruent to arc AD; so this is congruent to this; and prove that triangle BCP, this triangle here, is congruent to triangle ADP.
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I am going to do a two-column proof with my statements and my reasons.
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Now, remember that, since we are trying to prove that two triangles are congruent...
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remember the unit where we had to use either Side-Side-Side, Side-Angle-Side, Angle-Side-Angle, or Angle-Angle-Side.
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We have to use one of these four theorems and postulate to prove that triangles are congruent.
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That means one corresponding side, another corresponding side, and then a third corresponding side.
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Those parts have to be congruent, and then we can prove that those triangles are congruent.
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The first statement is going to be arc BC being congruent to arc AD; the reason for that is "given."
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And the next step: if those arcs are congruent, then I know that these chords are congruent.
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And these chords are parts of the triangles; so I am going to say that BC is congruent to AD.
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And then, I am making them segments of the triangle; what is my reason?
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It is the theorem that says that, if two arcs of the same circle are congruent, then corresponding chords are congruent.
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I am just giving my reason there: this theorem that says that, if two arcs from the same circle (or from congruent circles)
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are the same--are congruent--then their corresponding chords are congruent--that theorem doesn't have a name.
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So, you just have to write it out; you can shorten it--I shortened it a little bit.
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And then, the third one: that gives us a side, and they are all using sides; now, what else can I say?
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I can say (here is a side; I am just going to put S there, so that way I know that I proved that one of the sides is congruent),
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from these triangles, that these angles are congruent; so angle BPC is congruent to angle APD.
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And my reason for that is "vertical angles are congruent"; any time that you have vertical angles, they are always congruent.
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So then, there is an angle there; and then what else?--I need one more.
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I have a side, and I have an angle; that means that I am not going to be using this one.
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And for the next one, if these chords were parallel, then I could say that this angle is congruent to this angle, because they are alternate interior angles.
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But I don't know that they are parallel, so I can't use that reason.
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Now, what I can say, though, is that this angle here, angle B, is an inscribed angle.
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I will just draw this so that it is easier to see; this is the inscribed angle, right here, with an intercepted arc CD.
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This angle right here is an inscribed angle intercepting this arc; now, this angle here, angle A, is also an inscribed angle,
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so this angle and this angle are both inscribed angles, intercepting the same arc.
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What do we know about two inscribed angles with the same intercepted arc? They are congruent.
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Just to make it easier to see, here is angle B, and here is angle A; this is the same, and this is the same, so they are congruent.
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This one and this one...angle CBP is congruent to angle DAP (I wrote B), and the reason for that: "Inscribed..."
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and again, this theorem doesn't have a name, so you just have to explain it "...angles with the same intercepted arc are congruent."
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So, here is another angle; now, since we have two angles, and we have a side, this can either be this one or this one.
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We have to see what the order is: is the side the included side from those?
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We have Angle-Angle-Side; it wouldn't be Angle-Side-Angle, so it would be this one right here that we are using.
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My fifth and final statement is going to be the statement right here: Triangle BCP is congruent to triangle ADP.
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What is the reason for that? Angle-Angle-Side.
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To review, the given was that this arc and this arc are congruent; since they are within the same circle,
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their corresponding chords will be congruent; and that is what we used for the first one right here; that is a side.
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Then, for the third one, we said that this angle and this angle are congruent, because they are vertical angles; and that is an angle, right there.
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And then, we said that this angle B and angle A are congruent, because they are both inscribed angles intersecting the same arc.
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It is like if this is, let's say, 80 degrees, the inscribed angle is half the intercepted arc; so if this is 80, then this has to be 40.
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Well, this is also an inscribed angle with that same arc; so if this is 80, then this has to be 40.
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So then, inscribed angles with the same intercepted arc are congruent, and that was the last piece that we needed to prove that the triangles are congruent.
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And the rule is Angle-Angle-Side; that is it for this example.
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And that is it for this lesson; thank you for watching Educator.com.