 Section 10.4 is all about inscribed angles, and 10.5 deals with tangents. Make sure you get this in-deer table of contents, and then we'll get started. To begin, we need to understand what an inscribed angle is. An inscribed angle is an angle that has its vertex on the circle, and the sides of the angle are contained in chords of the circle. So if you recall, a chord is a line segment that has its endpoints on the circle. So here, angle ABC is an inscribed angle. That red arc, arc AC, is an intercepted arc. In other words, the inscribed angle cuts off a portion of the circle. It intercepts it, and that portion of the circle in the interior of the inscribed angle is arc AC. And so that brings us to our first theorem, which states if we have an inscribed angle, then the angle's measure is half the measure of its intercepted arc. So if we have our circle in an inscribed angle, let's say it's 50 degrees, that tells me that the arc that is intercepted is 100 degrees. There's this relationship that the angle is always half as big, or the arc is always twice as big. Here's another theorem for us. If two inscribed angles intercept congruent arcs, then the angles are congruent too. So if we look back at this original drawing, let's say there's another inscribed angle that intercepts the same amount of circle. So the black angle cuts off 100 degrees. The green angle also cuts off 100 degrees. That means both angles are congruent. Now similarly, if we have these two angles and they intercept congruent arcs, maybe not the same arc, but the same amount of arc, that means those two angles are congruent. So let's take a look at an example here. Let's say we have a circle, and we have an inscribed angle, and we also have a central angle. Let's say the central angle is 40 degrees. I know from previous lessons that the central angle and the intercepted arc are congruent. From this lesson now, the inscribed angle is going to be half as big. Let's take a look at this next example. Here we have arc AD is given as 4x degrees. We see ABD is 36 degrees. If you look at angle ACD, it is also an inscribed angle, and it intercepts the same amount of circle as angle ABD. So therefore, both those angles are 36 degrees, and arc AD must be 72 degrees. So therefore, we would get a value of x is 18. For another inscribed angle theorem, this states if an inscribed angle intercepts a semicircle, then the angle is a right angle. So if we have, let's take a look at that kind of bottom corner angle, we see it is an inscribed angle, and it intercepts a full semicircle. Well, semicircle is 180 degrees, and so therefore the angle must be half as big. Then our last theorem dealing with inscribed angles, if we have a quadrilateral inscribed in a circle, then opposite angles are supplementary. So inscribed means the quadrilateral is drawn entirely in the circle, and its vertices are on the circle. So let's call this LMNP. If you look at angle L, I see angle L intercepts arc MNP. The angle opposite, L, is M, and angle N intercepts PLM. Well, those two arcs, the red and the blue, make one full circle. And so therefore, angle L and angle N, their measures have to sum up to 180, and similarly angle P and angle M also add up to 180. Next example, let's say we have circle P, we have a diameter, Tu, and we've got these points, V and S. And let's say we're given that arc VU is congruent to Su, angle UTS is X plus 9, and angle SUT is 2X plus 16. Let's say we want to solve for all of the missing angles. Well, we know we've got these congruent arcs. And those congruent arcs, both of those arcs are intercepted by either STU or UTV. So therefore, those two angles are congruent. We have a semi-circle up the middle, and so that means angle S must be 90 degrees because it intercepts a semi-circle. Similarly, angle V is 90 degrees. And so if you just look at triangle TUS, we know all three angles. So therefore, we can set them equal to 180 and then solve for X. So now that we have got a value of X, we can sub that in and find the rest of the angles in the triangles. You know this is 34, TUS is 56, VTU is 34. And then looking at triangle VTU, I know angle VUT must also be 56. One last example for this part. We've got quadrilateral ABCD, and we want to solve for the unknown variables. Well, we know with an inscribed quadrilateral opposite angles are congruent. So B and D add up to 180. And likewise A and C add up to 180 as well. Next, a little bit more vocabulary from section 10.5. So quick review, we all know that segment AD is a radius. Segment BD is a chord. Remember a chord is any line segment that has its endpoints on the circle. So now look at line BC. Line BC is called a secant. Now a secant is a line that intersects a circle in two points. Secants and chords are very similar ideas. A chord, however, is a line segment that intersects a circle in two points. Whereas a secant is a line that intersects a circle in two points. Now look at line DE, that's called a tangent line. And a tangent is a line that intersects the circle in only one point. A secant is two points, and a tangent is one point. So that point is called the point of tangency. One thing about the point of tangency is that the tangent and the radius are always going to be perpendicular, which means angle ADE is 90 degrees. So let's look at this example problem. Let's say circle A has a radius of x units long, and B is a tangent point, and we want to solve for x. Well we know if we have a tangent point, that means angle ABC is a 90 degree angle. Then take a look, AB is a radius, AD is also a radius. And so that means the entire length AC is x plus 8. And so therefore triangle ABC is a right triangle. So we can solve it using the good old Pythagorean theorem. Now make sure you foil out x plus 8. Then I'll subtract x squared from both sides. Subtract 64, and finally divide by 16. Next we have a theorem about tangents. And it states if two segments from the same exterior point are tangent to a circle, then those two segments are congruent. So first let's figure out what this means. We have a circle, and we have a point exterior to the circle. If we draw the two segments so that they're tangent to the circle, what this theorem says is that those two lines, those two line segments are congruent. This is actually a pretty easy theorem to prove. All you have to do is let's say draw in the radii that connect to the points of tangency. Remember tangents and radii are always perpendicular. And then if you draw the distance or the segment between the center of the circle and point A, let's name the rest of those. Right now we can prove that the two triangles ABD and ACD are congruent, right, by HL. And therefore, segments AB and AC must also be congruent by the CPCTC. So there we go, we've got that tangents theorem. I sincerely apologize for the quality of this drawing. Let's say that we have these two circles and these segments and we've got tangents. Sorry about that circle, but let's pretend. And then let's say we know these segment lengths are given. We've got y minus 5, 10, y, and x plus 4. And so if you just look at that small circle, that small horribly drawn circle, we know that these two segments are tangent from the same exterior point. And so that means right off the bat y is 10. Then if you ignore that small poorly drawn circle and just look at these two tangent lines, they're also tangent from the same exterior point. So the blue lines must also be congruent in length. That top side, we know what y is, y is 10, and so therefore the entire length there is 15. And so likewise the bottom length must equal 15. Let's take a look at one more example here. Let's say we have a circle inscribed in a triangle, and the triangle's perimeter is 30 units, and we have to find x. First off, if we have a circle that's inscribed in a triangle, that means we create these three points of tangency. And so since we have exterior points, we know that if that little segment is 7, so is the one right below it. Likewise we have these two segments that are 5 units long, and these two segments that are x units long. And so since the perimeter is 30, we can just add up all of the line segments. 5 plus 7 plus x plus 5 plus x plus 7 must equal 30. Then we'll go through a cleanup session. We see 2x plus 24 is 30, which means x is 3. Thanks for watching.