 Hi, I'm Zor. Welcome to Unizor education. I would like to continue our discussion about circles. And today I will go through another set of mini theorems. Mini because they're really very simple. However, again, the training you give to your logic, to your analytical abilities while trying to prove these theorems is really valuable. So I do recommend you to go through these theorems first from the notes and try to prove it yourself without actually reading my proof, which is also in the notes. And I will just present the same thing in a conversational kind of a mode. Maybe I will come up with something new, which is not exactly in the notes, so just bear with me. Okay, start from the beginning. Some of opposite angles in the convex quadrangle inscribed into a circle is equal to 180 degrees. Okay, that doesn't seem to be difficult. So you have this circle quadrangle inscribed in it, a convex quadrangle. Okay, now we have to prove that some of opposite angles is equal to 180 degrees. Well, as you know, any inscribed angle in the circle and angle ABC, for instance, is inscribed is equal to half of the central angle, which is supported by the same arc. Now, in this case, arc is this. It supports the angle ABC. So what's the central angle? It's this one. This is the central angle, which corresponds to the same arc. Now, opposite to this angle is this, ADC. Now, it's supported by this arc and central angle, which is supported by the same arc is this one. Now, it is obvious that these two arcs are actually making a full circle and these two central angles in sum give you 360 degrees. So these two angles in sum, since each of them is half of the corresponding angle, this is half of this angle and this is half of this central angle, so their sum is 180 degrees. Now, analogously, sum of these two angles also is 180 degrees because arcs are this one and this one, and again, they together give you the full circle. So this is simple. Now, next theorem is in verse and it says if sum of the opposite angles of a convex quadrangle is equal to 180 degrees, then it can be inscribed into a circle, which means that if you will take, let's say, any three vertices of this quadrangle and have a circle around these three points, the force point, the force vertex will be on that circle. So that's what I'm going to prove. So I have a quadrangle and I know that the sum of opposite angles is 180 degrees. Now, how can I prove that you can draw a circle around all four points? Okay, let's assume that you cannot. So let's assume that these three points, which you are using, are not actually making a circle which is going through the force point, which means the force point is either outside or inside of that circle. So let's consider two cases. For instance, the force point is outside. So this is our quadrangle. Now, I know that some of the opposite angles is 180 degrees and I would like to prove that this is impossible. All right, let's connect this line with... Now, if D is outside of the circle, then there is one point where the side A D is crossing the circle and that's the point E. Now, in case it might actually be a tangent, then the other side would be crossing. So this is the crossing and I connect the opposite vertex with the point where A D is intersecting the circle. Now, let's consider this quadrangle, A B C E. Now, we know that this particular quadrangle is inscribed into the circle, which means the sum of... Actually, I don't even need this one. So we know that the sum of opposite angles is equal to 180 degrees, which means this and this angle together gives 180 degrees. But now, let's consider this triangle, E C D. This is exterior angle, that's why it's equal to the sum of this plus this. So it's definitely greater than this angle. So if this plus this is 180 and this one is greater than this, then B plus G angles are supposed to be less than 180, which contradicts our initial assumption that the whole quadrangle has opposite angles, sum of opposite angles equal to 180 degrees. So the point D cannot be outside of a circle. Now, let's consider another case. For instance, it's inside the circle. Is that possible? Okay, so let's consider D is here. Again, let's just continue the line A G until it meets point E and connect it again with C. Now, what do we know now? We know that again, B plus angle E, since this is an inscribed, A B C E is an inscribed quadrangle, then angle B plus angle E give you 180 degrees. But now, considering triangle G C E, the angle D is exterior, which means it's greater than the angle E, which means that the sum of D plus B should be greater than 180. So again, we came to a contradiction. So both our assumptions that the D is outside or the D is inside a circle while still having sum of 180 degree lead to contradiction, which means D cannot be neither outside nor inside, it's on the circle. That ends the proof. Alright, next. Well, next are two very simple consequences from this theorem. The consequence number one, if you have a parallelogram, can the parallelogram be inscribed into a circle? Well, we know that the opposite angles are equal in any parallelogram. Now, we also know that if it's inscribed into a circle, the sum of opposite angles must be equal only to 180 degree, which means each one of them is 90. So the theorem is that among all the parallelograms, only rectangle can be inscribed into a circle, because with a rectangle, we have two opposite angles equal to 90 degree and equal to themselves. And that's why the sum is equal to 180 and this is a characteristic property of the inscribed quadrangle. The sum of opposite angles must be equal to 180. So among all the parallelograms, only the rectangle, the one with 90 degrees interior angles, can be inscribed. Now, next is trapezoids. Now, we have trapezoid. So these two are parallel. Now, we know that some of these two angles, since these are parallel and this is transversal, these angles are called one-sided interior, right? Their sum is equal to 180 degree of all this. Now, we also know that the sum of the opposite angles of the inscribed quadrangle must be 180. So this plus this is supposed to be 180. Now, let's just think about this plus this is 180 and this plus this is 180. What does it mean? It means that these two angles are congruent to each other. Similarly, these two angles must be congruent to each other. So only trapezoids with two angles at each base congruent to each other can be inscribed into a circle. Now, and what are these trapezoids with angles at the base congruent to each other? Well, this is the trapezoid when two sides are equal obviously to each other. So the trapezoid which can be inscribed must have congruent sides which are not parallel to each other. By the way, there is some kind of a duality in in calling a parallelogram also a trapezoid. So in this case sides are equal in lengths to sides of the parallelogram. However, this is not sufficient to be inscribed into a circle because as we were saying parallelograms which are not rectangles do not have some of these angles equal to 180 degree. So we are considering trapezoids only those which have not only equal in lengths the sides but also non-parallel to each other. So this is not the trapezoid with equal sides which we are considering in this case. So only trapezoids with equal and non-parallel sides can be inscribed into a circle. So these are very, very simple consequences of this characteristic property of the inscribed quadrangle to have opposite angles equal to sum of opposite angles equal to 180 degree. A quadrangle that circumscribes a circle that is all four its sides are tangent to a circle has sums of opposite sides equal to each other. So this is instead of inscribing we are circumscribing. So we have a circle inscribed into a quadrangle or a quadrangle circumscribed around a circle. Now this particular theorem says that the sum of the opposite sides is the same. Sum of AD plus DC is equal to sum of AD plus DC. How can we prove it? Actually it's very easy. We know the theorem about a point and two tangents from it. And the theorem says that these tangents are supposed to be equal in lengths. Now how can it be proven? Well, very easily. We know that the radius to the point of tangency is perpendicular to the tangent here and here. So these are two right triangles with equal tangency and shared hypotenuse. So they are equal to each other which means this is supposed to be equal to this. I just remind you very simple proof. But in theory we don't really have to prove it because it's kind of a known fact and it was proven before in one of the lectures. So how can I use this particular property? Well, let's just think about it this way. K, L, M, N. This is tangent which means this particular piece of that tangent from A to K is congruent to this from A to M because these are two tangents from one point. Similarly, these two are congruent. These two are congruent. And these two are congruent to each other. Now, if you will add this, this, this and this one, two, one, two, three and four you will get exactly the same segments as this, this, this and this. So that's why the sum of these is equal to the sum of these. You have exactly the same components. One strike, two strikes, three strikes and a V. Here also one strike, two strikes, three strikes and a V. Right? So that's why the sum of opposite sides of the circumscribing the quadrangle is actually equal to each other. Okay, next. Okay, this might be a little longer. About four special points in any triangle. Triangle has a certain number of really very, very special points. We did talk about this many times before. I would like to devote a certain amount of time in this lecture to kind of summarize all these properties. So what are these four remarkable points which you can find in a triangle? Point number one is the intersection of perpendicular bisectors to the sides. So you have three sides and you have perpendicular bisectors to two of them. Now, first of all, why does this intersection exist? Well, obviously because these are not parallel, which means perpendicular to non-parallel sides are also non-parallel to each other and that's why they are intersecting. They are not intersecting necessarily inside a triangle. They can actually intersect outside. If you have something like this, you have perpendicular bisector one and two, the intersection is outside. So there is an intersection. That's number one. That's the property. Property number two is a center of a circle which circumscribes this triangle. Why? Because the perpendicular bisector is a locus of points equidistant from two points. So this perpendicular bisector is locus of points equidistant from this from B and C. So every point on this perpendicular bisector is equidistant from B and C. Now, this perpendicular bisector to the side AC is locus of points which are equidistant from A and C. So the intersection of these two is equidistant from B and C and C and A, which means it's equidistant to all three of them, which means this is the point where you can put the center of a circle which circumscribes this triangle with this piece as a radius because these are all three the same. So the point which is a center of circumscribed circle can be found by using these perpendicular bisectors. Now, the question is why does the third perpendicular bisector, which is this one, is crossing these two in exactly the same point? Well, this doesn't seem to be as trivial as just the one, the statement which I made before about having one point equidistant from the all three. And here is the logical conclusion which you might actually see. The first and the second perpendicular bisectors, bisector to BC and bisector to AC, are crossing in certain point and this is the only point which is equidistant from these three actually, from these three vertices. Now, if you consider that this third perpendicular bisector is not actually going through this particular point, let's say it's going through some other point. So what do we see right now? Well, this is the point which is also equidistant from, in this case, B and A, from A and C, which means it's also equidistant from all three points, which means it's also supposed to be on this line, on the first perpendicular bisector. So it looks like the first perpendicular bisector crosses the second perpendicular bisector first in the point where we were talking about the first time and the second it should actually go through this point as well. Since two different lines cannot intersect in more than one point and this is probably the most important axiom which we are using here. Two lines, non-parallel to each other, can intersect only in one point and that's why the third perpendicular bisector should actually go through the same point which we have already received by intersecting to other perpendicular bisectors because otherwise it would mean that the lines are intersecting in more than one point, which is not good. Okay, so this is my first remarkable and very interesting and special point, a center of a circle which you can circumscribe around the triangle. It's on the crossing of all three perpendicular bisectors to the sides. So, that's circumscribed. Now, how about inscribed circle? Again, triangle. Now, if you have an inscribed, now what's the characteristic property of inscribed? It means that every side is a tangent, which means if I will put radius into each one of these three points of tangency, these are three radiuses and these are right angles, right? And this must be an angle bisector. Why? Because you have this triangle and this triangle. You have radiuses which are the same and you have shared hypotenuse. These are right triangles. These casualties are congruent to each other and hypotenuse is shared, which means they are congruent and that means these two angles are supposed to be congruent as well. So, the center lies on the bisector of the angle. And obviously it lies on the angle bisector of this one and this one. So, if you connect a center with all three vertices, you will get three angle bisectors. So, the second important, special, remarkable, whatever point in a triangle is the point where three angle bisectors are crossing each other. And this is the center of an inscribed circle. So, perpendicular bisectors to sides give you the center of circumscribed circle. Bisectors of the angles give you the center of inscribed circle. Next, next are altitudes. Now, altitudes are also crossing in one single point. And here is how to prove it really very easy. This is an altitude, this is an altitude and this is an altitude. I would like to prove that they cross each other in one point. All three lines are crossing in one point, which is the center of altitude. Now, how can we prove it? Here it is. From each vertex we draw a line parallel to opposite side. And from this we will draw parallel to this, from this parallel to this and from this parallel to this. Now, consider these triangles, one and three around it. They are obviously congruent to each other. Why? Because these are parallel lines. So, this is the parallelogram. Diagonal always divides the parallelogram into two congruent triangles. This is equal to this, this is equal to this and this is equal to this. For obvious reasons. This is parallelogram. So, these are two opposite sides of the parallelogram and this is parallelogram. So, this side is congruent to this one. That's why they are congruent to each other. Same thing here and here. But now, let's talk about our altitudes. They are perpendicular to the size of the big triangle. So, this altitude towards this side of a main triangle is a perpendicular bisector to this side of a bigger triangle. Same thing here. This altitude to this side of a main triangle is a perpendicular bisector of this side of a bigger triangle. And same thing here. So, our three altitudes of our original triangle become perpendicular bisectors of the sides of this bigger triangle. And we have already proven that the perpendicular bisectors of the sides are crossing in the same point, which is the point of the center of circumscribed circle. So, basically from here, these distances are the same. And this is the center of the circumscribed about the bigger triangle. Okay, so that's how we prove the intersection of altitudes is one and only point, that all three altitudes are crossing in the same point. Now, and the last remarkable point. So, we have perpendicular bisectors of sides. We have bisectors of angles. We have altitudes. They're all intersecting in one point. By the way, different point. I mean, it's not the same point for different triangles, it's obviously different points. And the last one is medians. Medians are also intersecting each other in one single point, which is, by the way, the center of gravity, but that's the physics we don't really touch this. So anyway, three medians, how can we prove that these three medians are intersecting in one point? Okay, let's consider these two medians and the middle of that line. So we have ADC, we have Mn, and I would like Mn, whatever, L, and I would like from point L, now Mn and L are all midpoints of the sides of this triangle. That's why Vm is a median and Cn is a median. Now, if I draw AL, AL also would be a median, but I don't want to draw AL right now. So let's just put the parallel line KL parallel to Cm, and another parallel line from M also parallel to Cm. Now, what do we see now? Consider triangle MbC. KL is the mid-segment because it's parallel to the base and it's dividing one side in half because L is half of the BC. That means that the other point where this mid-segment is intersecting the other side is also divided in half. So these two segments are congruent, Bk and Km, since these are congruent to each other and these are congruent to each other. Now, in this triangle ANC, same thing actually, AM and MC have the same lengths and Mp is drawn parallel to Cm. So if it divides by half one side of a triangle and it parallel to the base, it divides another side also. So these two are congruent to each other. Now, and why are they all congruent to each other? Well, because this is a midpoint, which means this actually cuts one quarter and one quarter, and this is actually one quarter and one quarter of the side AB. So each one of these segments is one quarter of the AB and they are equal to each other, this and this and this and this. They are all supposed to have the same lengths. Now, let's consider triangle Bpm, this one, Bpm. As you see, we have a very similar situation. We have equal segments on one side of this triangle and lines parallel to the base. The base is Pm right now. Now that means that another side is also divided equally. We actually did go through this theorem long time ago. If you have an angle and you have equal lengths segments and parallel lines, then these segments also would be equal in lengths among themselves. It's very easy to prove by drawing parallel lines here, and basically these triangles are congruent to each other. Okay, so what we have proven is that this particular point, which is a point of intersection of two mediums, is cutting this medium Bm in a ratio of two to one. So one third of Bm is where this point is located. Now, if I draw another medium, it must also be catching this piece from the Bm, which is equal to one third. It's exactly the same type of consideration, just reverse parallel lines. Instead of parallel lines to this, I would probably should put parallel line to this and to this, same for different segments, and that's why the same three segments are on this side. So in any case, whenever I have one single medium Bm, another medium cuts one third of it in this particular place, between the point of intersection and the side where it falls. And that means that the second medium, the third actual medium is supposed to also go through this point because it must cut one third. So incidentally, we have proven not only that three mediums are intersecting in the same point, but also that this point where they are intersecting is cutting one third of the medium towards the side where it falls. So this is one third of the medium, this is one third of the medium, and this is one third of the medium. So this is a property which we were using, we proved it and then we were using to prove that all three mediums are intersecting in one point. And as I was saying, this point of intersection of mediums is the center of gravity, which means that you will make a triangle out of, let's say plywood or something, draw three mediums and support it from this point where the mediums are intersecting, it's supposed to be in equilibrium. Okay, so four major points in a triangle, intersection of perpendicular bisectors to the sides, intersection of bisectors of angles, intersection of altitudes and intersection of mediums. All these intersections are points where all three components, all three mediums or all three altitudes, bisectors, et cetera, but they're crossing. And they are remarkable in a way because all these points are signifying something. The center, let's say, of a circumscribed circle is the crossing of perpendicular bisectors, et cetera. All right, next. Given two points on the plane K and L, okay, consider all straight lines that contain point K. Let's say this line, this line, and this line, all of them. And perpendicular to these lines from the point L. So perpendicular to this one is this, perpendicular to this one is this, and perpendicular to this one is this. So question is, points where these perpendiculars meet the corresponding lines, this point, this point, this point, and all other points, wherever and however we draw the line through K. Why perform a circle with a KL as a diameter? KL as a diameter. And they are all on the circle around this diameter. Okay, question is why? How can we prove it? Well, actually the proof is extremely easy because you see, these are all right angles, right? Now, you remember that right angle is always supported by the diameter of a circle if it's inscribed. So that alone actually should give you this type of clue. But at the same time it's very easy to prove that you see these right triangles, they're all right triangles, and right triangles we have already proven many times has the property that this particular point, the midpoint of a hypotenuse, is equidistant from all three. Which means this distance is always equal to this and always equal to that. If you don't remember you can actually try to look at this and prove it yourself actually, it's very easy. So basically it means that the midpoint of this KL is equidistant from this as well and this as well and this because all these distances are equal to OK and OL. So they're all supposed to be on the same circle. Given a circle with a center L and point K inside it, consider all chords that contain the point K, all different chords. Midpoints of these chords lie in a circle with a segment KL as a diameter. Midpoint, right, but you know again that the midpoint connected to the center is always perpendicular to this connection. Midpoint of another chord connected to a center is again perpendicular, connection radius is perpendicular to the chord and here they're same. So basically we have exactly the same situation as before. All these triangles, this, this and this, they're all having the same hypotenuse. Which means again, these are all right triangles and that means that they're all supposed to be on a circle which is using hypotenuse as a diameter. It's exactly the same as the previous problem. Nothing new. All right, number eight. Given two circles tangent to each other and whenever you see the two circles tangent to each other most likely you will need to connect the centers and you know that the center line actually goes through the point of tangency. I'm sure it's used here. Any second that contains a point of tangency forms two chords. All right, so let's make two chords. Second makes two chords. Okay. Proof that two central angles supported by the arcs that correspond to these two chords which means this central angle and this central angle are congruent to each other. All right, well, that's actually easy. Look at it this way. Now, since this is a line, one line, it's not angular. It's one line between two different centers and it goes through this point. And this is the line. So these are vertical angles, right? These are vertical angles. Now, these triangles are esosceles and this one because these are radiuses, right? So some of the angles of any triangle is 180 degrees which means that this angle at the top of this triangle is equal to 180 minus this and this but these are congruent to each other because it's an esosceles triangle. So this one is equal to 180 minus two of these. That's called alpha. And this one will be called beta. So beta equal to 180 minus two alpha in this case and similar in this case because it's also an esosceles triangle, these two are equal to alpha and the beta is equal to 180 minus two alpha. So that's what makes these angles the same. Beta here and beta there. Okay. You know what I'm thinking about right now? After you have gone through all these proofs which I have presented here if you did not do it yourself go through the notes on the website and try to not to read my proofs over there just read the theorem itself and try to prove it yourself. I think that would be a great exercise. And again, don't forget unizord.com is the source of knowledge especially mathematics, advanced level of mathematics for high schools. Use it as much as you can and don't forget that none of the information which is presented on this side would actually be very useful in the future in your professional life. It's the skills of analytical thinking and logic which are supposed to be developed by solving these problems and going through these theorems and their proofs. That's what's supposed to be used in the future. And parents, teachers, supervisors please take a look at this website especially how to section which explains that you can be in control of the education of your students and basically they will be enrolled to certain programs. They can go through exams. You can check the score on the exams and mark certain programs as completed or not completed basically. Thank you very much and until the next time.