 Hi, I'm Zor. Welcome to Unisor Education. I'm continuing talking about circles and proving certain theorems about circles. You might have noticed I called this particular lecture theorem number one, not mini theorem something. Now why? Because primarily mini theorems are those which can be just proven in three, four statements. These are just a little bit more complex, not by much, just a little bit. Some of them which I will present here will seem to be a little bit simpler, but there is one or two theorems which might actually be a little bit more complex and that's why I decided to call this theorem as a more kind of a complicated or sophisticated, if you wish, theorem. So anyway, let me get to these theorems. Number one, given two chords of equal lengths crossing each other in the same circle and divided by a point of intersection in two unequal segments each, short and long. So we have two chords of equal lengths. Now that's not exactly equal lengths. Let's assume these will be equal lengths, hopefully. All right. Prove that short segment of one chord is congruent to a short segment of another. So we have to prove that these short segments are congruent and the long segments are congruent. Provided the lengths of chords is the same. Well, that's actually kind of a mini theorem. I probably should have put it in a different category, but still. Now, obviously, if you have chords, the best way is to draw perpendicular to these chords from the center. And now, obviously, these triangles will be congruent because the radiuses are hypotenuses and these two, casualty a, m and g, are half of the chords and chords are equal in lengths, so the halves are equal as well. Now, similarly, if you connect these two guys, if you can consider these two triangles, m p o and m p o, these two small triangles, they are also congruent because, as we know, o m and o n are congruent. These are part of the congruent bigger triangles. And o p is shared and they are right triangles, so they are congruent as well, which means these little segments m p and m p are congruent. So what remains is basically c p and b p. These two must be congruent because these two parts of the congruent chords have the same lengths. So that's it. Yeah, this actually doesn't deserve to be called a theorem. This is really a mini-theorem. But still, there will be one or two little more complicated problems here. Okay, number two. Two circles intersect at points a and b. A second is drawn from a intersecting at c and d. Okay. So you have two circles, not necessarily the same radius. Intersection points are a and b. We draw a second which intersects at c and d. Prove that measure of c b d, c b d. Prove that measure of angle c b d doesn't depend on the position of the second c d as long as it goes through the point a. Okay. All right, consider this triangle b c d. Angle b c d is supported by this arc. And no matter how we position the point c, let's say it's this way, c prime, d prime, and connect it again the same way. So no matter how we position the point c, this angle will still be the same and it will be supported by the same arc. Because this is an inscribed angle and it's always equal half of the central angle which is supported by the same arc. So this angle is constant. Same thing with this one. This is supported by this arc and it's always half of the central angle which is also supported by this arc. Which means these two, both these two angles are constant regardless of whether we draw second this way or this way. Since the sum of the angles in a triangle is always a hundred and eighty degree, what implies is that this angle which is equal a hundred and eighty minus this and minus this is also constant regardless of the position of the second. That's it. Two circles with centers at points k and l are tangent to each other point m. Again two circles but in this case they are tangent to each other. So this is k, this is l and this is m and as you know the tangent, the line of between the centers always passing through the point of tangency between two circles. They're tangent. Two seconds intersecting both circles are drawn through the point m. Okay, one is this way from a through m to b and another from c through m to d. Okay, prove that quarts a, c in the first and b, d in the second circle are parallel to each other. So these two lines, these two quarts are parallel to each other. Okay, how can we prove that? Well, since this is the point of tangency, we can always consider the tangent line which is common for both and it's perpendicular to the center line. So let's, now let's just think about what can be actually said about all these angles. Now these are obviously vertical and therefore congruent to each other as well as these two angles. Okay, let's give it. What else? Okay, I do remember that angles formed by, let's just consider these triangles. Since a k m is an isosceles triangle, these are two bases, base angles. A m is a basis, k a and k m are radiuses. They are congruent to each other. Similarly, this triangle is also l m b is also isosceles. Therefore, these angles are the same. So as a result, we have that these two angles must be the same because this is 180 minus these two and this one is also 180 minus these two. Since these are vertical, these angles are equal. Now, this is a central angle and this is a central angle and this one is an inscribed angle which is supported by the same arc. So a m is an arc, so a c m angle equals half of this and b d m inscribed angle equals half of this central angle. Which means a c d and b d c angles are congruent to each other and that's why considering c d is a transversal and these a c and b d are two lines, angles which I was just talking about are interior alternate. There are interior between the lines and they are alternate because they are on both sides of the transversal. So again, lines are a c and b g. These are lines which we have to prove the parallelism of. The transversal is c g and these two angles are alternate interior angles. And since they are congruent to each other, the lines are parallel as we know from the theory of the parallel lines and transversal. So we don't really need this, common tangent. Everything is just proved from congruence of the central angles, therefore congruence of inscribed angles. You see, I was actually thinking about this a little bit, so that's why this is a theorem and not a minis theorem, but I don't have to think at all. Let's move on. Two circles with centers in points k and l are tangent to each other, so it looks like it's the same kind of a feature. k and l and they go through point m. And probably we need this as well as usually. If two circles are tangent, most likely the central line will be needed. A second intersecting both circles is drawn from a through m to point b in the second one. Alright, so we have a second from a. We draw two tangents, here tangent a and here tangent b. And we have to prove that they are parallel. Well, let's do exactly the same thing. So we connect these two points. So we definitely know that these angles, we have proven in the previous theorem that these two angles are congruent to each other. It follows from the congruency of these vertical angles and the fact that these are isosceles triangles. Alright, so we have these angles equal to each other. Which proves what? Which proves basically that a, k and l, b are parallel to each other. Because a, k and l, b are two lines, k, l is a transversal. And since these angles, which are again alternate interior, are equal to each other, that proves the parallelism of the lines. But now these lines, these two radiuses, are perpendicular to two tangents. Since these are parallel, then the perpendicular to parallel lines are parallel among themselves. So that proves the parallelism of two tangents. Okay, next one. If in any triangle, basis of three altitudes are connected to form a new triangle, then these altitudes will be angle bisectors in the new triangle. I see. That's actually a very interesting problem. And probably that's why I called the whole lecture theorems rather than mini theorems. Because this seems to be just a little bit more complicated. So you have a triangle and you have three altitudes. Here's the question. What happens if I connect the basis of these altitudes? The theorem states that in this new triangle, let me put letters on it, triangle A, B, C, basis R, M, and P. So in the new triangle, M and P, the old altitudes will be angle bisectors. Well, again, this is kind of unusual and I would say unexpected theorem. And that's what makes it actually very interesting. Personally, when I first time saw this type of a theorem, I was a little bit surprised. Well, it's interesting. If you connect them, if you connect the basis, it will be a new triangle and the altitudes will be angle bisectors. So you do not expect this type of a theorem. That's why it makes it really interesting and interesting to prove, actually. All right, so how can we prove that? Let's think about this way. You know that if you draw a circle around right triangle, then the center of this circle will be the midpoint of a hypotenuse and the hypotenuse itself will be a diameter. And any right triangle which shares the same hypotenuse would be inscribed in this circle. So hypotenuse defines basically, I mean the center of the hypotenuse and a circle around it using this as a diameter defines the locus of all points which if connected to these two ends would form right triangle. So all right triangles which share the same hypotenuse lie on the same circle. Now we definitely know this and the question is how can we use it? Well, simply consider AC as a hypotenuse and two triangles, AMC. It's a right angle, right? It's an altitude. And APC. So this is a circle. If we will draw it and both P and M will lie on this circle. Well, let me just use maybe almost some other color. I'll try to do it as accurately as I can so it will be visible. All right, so this is a circle, in this case it's half a circle where AC is a diameter. That's why AMC triangle will be inscribed into the circle as well as APC. And now let's consider angles which are supported by the same arc. Well, let's look at the arc and P. Now obviously inscribed angle PAM, this one, and PCM are supported by the same arc. Again, this angle and this angle both are supported by this arc. Now both are inscribed angles. Both are equal to half of the central angle which is supported by the same arc which means they are congruent to each other. Okay, we got that. Now let me wipe this out, raising this old circle so it doesn't really bother us anymore because we will draw another one. We have three altitudes, right? So we can have three different hypotenuses. Now let's consider the side BC as a hypotenuse and triangle BNC as well as BPC sharing this as a hypotenuse. So BPN and C will be right in the same circle, right? Now let's consider these angles. Again, sharing exactly the same arc. BCP is supported by this arc and BNP also supported by the same arc which means this is the same angle. So BCP angle and BNP angle are congruent to each other. And finally, the third circle, erase this one and use AB as a hypotenuse. So triangle BMA, right triangle, and BNA would be inscribed into this circle. So circle will look like this. Now in this circle, let's look at this particular arc, BM. This angle BNM is inscribed and supported by this and BAM is also inscribed into the same circle and supported by the same arc. So they are congruent to each other. So this one and this one. And that actually concludes the proof that these two angles are congruent to each other because this one and this congruent from the first circle, if you remember, then this and this from the second circle and this and this from the third circle. So these are congruent to each other. That's why BM is a hypotenuse. Now, what I actually can do is I can repeat exactly the same logic for these two angles and these two angles. I mean, I could have actually done it in the first place to the first circle not only to specify these two, for instance, angles congruent to each other but others as well. But that would be a little bit too much for one circle. It will be too many different angles. So I decided to concentrate only on these two angles. And this is a proof. But the proof is actually very symmetrical, if you notice. One circle, another circle, and third circle. So we can actually do exactly the same thing for any pair of angles. So that proves that the altitude is a bisector of a triangle formed by the basis of three altitudes. Interesting and quite frankly, again, for myself it was an unexpected, I would say, theorem. Alright, given an equilateral triangle, ABC inscribed into a circle. So there is a circle and there is a triangle ABC inscribed into it. Any point M on a circle, all three vertices to this triangle. So we take any point M here and we connect it with three vertices. So we have this, this, and this. Alright. Okay, we have to prove some of two shorter connections, AM plus MC is equal to the long one, BM. So again, triangle ABC is equilateral. Actually, this center point supposed to be relatively in the center of it, but that doesn't really, we don't need really the center point. So we have an equilateral triangle and we take any point on, let's say, on this particular arc between A and C and connect it with three vertices. And we have to prove that the sum of two shorter connections is equal to one longer one. Okay. So what do we know about this? We know that this is equilateral triangle, which means that these angles are all 60 degrees, right? Since it's equilateral triangle, they're all 60 degrees. In addition, since angle BMC is supported by the same arc as BAC, this is also 60 degrees, as well as this is 60 degrees. So these are all angles equal to 60 degrees. In this case, BMA and BCA supported by the same arc. Now, basically, if we would like to prove that this big segment is equal to sum of these two, what I think makes sense is basically to have another equilateral triangle from the point C and M and put the third vertex on the BM. So what I will do is I will draw a line here such that this angle is also 60 degrees, which means in this one will also be 60 degrees, which means this is equilateral triangle. And this particular segment, CM, and let's call this one P, would be equal to PM. Since PCM is equilateral triangle, all angles are 60 degrees, so it's equilateral. Now, we know this and so we took this piece CM and put it on this particular segment as an MP. What remains to be seen that the remaining lengths, which is BP, is congruent to AM. Now, to do this, we probably have to include it in some kind of a triangle. So let's think about it. We can consider these two triangles and prove their congruence. VPC and ACM, let me write it down. VPC and triangle AMC. Now, let's think about these two triangles. VC is obviously congruent to AC because ABC is equilateral triangle, right? So we have VC equals to PC. That's number one. What else did we know about this? We also know that PC is congruent to AMC. PC is equal to AMC. So this is second side. So we have one side congruent to this and we have another side congruent to this. So the only question actually remains whether the angle between these two, BC and PC, BC and PC, this angle, is congruent to angle between PC and AMC, which is this angle, AC and AMC. So in this triangle, we have side equal to side, another side equal to another side. And the only thing which remains is this angle against this angle. Now, well, actually it's very easy to prove. Why? Because this angle, angle BCP equals 60 degrees minus this one, PCA, minus angle PCA. Now, angle ACM is equal to, again, 60 degrees from the PCM, 60 degrees, minus, again, exactly the same angle PCA. So that's why angle BCP and ACM, this one and this one, are congruent to each other. So we have side, angle, side congruent to side, angle, side. So triangles are congruent, which means AM and BP are congruent to each other. And that's exactly what remains to be proven, because now AM being here and MC being here. Now they are making up one connection from B to M. So the segment BM is equal to BP, which is congruent to AM, plus PM, which is congruent to CM. And that proves the whole theory. That's it. That was the last theorem I wanted to prove. And I hope you agree with me that these are slightly more complex theorems than I used to do before in the category of mini theorems. But they're not really very, very complex. Just maybe a little bit. So thanks very much. And don't forget Unisor.com is the site for you and for parents and the teachers and supervisors who would like to control the educational process of their students. Good luck. Thank you very much.