 Hello friends, welcome to another session on Gems of Geometry. So, we are going to discuss another very interesting theorem here. The theorem says let triangles be constructed externally on the sides of an arbitrary, arbitrary triangle, so that the sum of the remote angles is 180 degrees, then the circumcircles of the three triangle meet at a point. So, let me explain this theorem to you. So, what is this saying? It is saying that let us draw an arbitrary triangle first of all and then you construct external triangles like that, like that on the three sides, on the three sides, but the condition is that the remote angles, what is the remote angle I will illustrate? So, let us say this is three triangles you are creating. So, this three angles alpha, beta and gamma, these are called remote angles. What are they called? Remote angles, let us say. So, that is what is meaning of a report angle in this case. It is not a standard, what do you say in standard or conventional way of identifying these angles or report angles, but just in this case remote meaning farthest away from the sides of the triangle, given triangle. So, alpha, beta, gamma, if alpha plus beta plus gamma is given as 180 degrees, this is given. Alpha plus beta plus gamma is 180 degrees, then they are saying that if you draw the circumcircule of these three triangles. So, they are going to meet at a point. So, I am just drawing a rough diagram. Anyway, the challenge is to construct this now. So, they are saying they will meet at, they are going to meet at one particular point. That is what we have to establish. So, understood. So, again there is a triangle ABC, this black one and then alpha, beta, gamma are the remote angles drawn or these are the triangles basically on the sides drawn externally and alpha, beta, gamma are the remote angles such that alpha plus beta plus gamma is 180 degrees. Then the three circumcircles of the three triangles which are drawn like these will meet at one point. That is what we are going to see. First of all, the challenge is to construct this, construct this particular thing because alpha plus beta plus gamma to be 180 degrees is the challenge in construction. So, what do we do? So, what I am going to do is first I will take a random triangle. So, let me take a random triangle. So, here is my point A. Let us say this is point V and let us say point C. And let us make, let me make this more arbitrary. So, how to make it more arbitrary? So, just change this orientation. Yeah, somewhat like that. And let me zoom in. Now, what I am going to do is the two triangles, the other two triangles on A C and A B. We can make those triangles very easily without having much of a constraint. So, let us say this is, this is one. So, one triangle I draw. Let me just change the color so that becomes little, which color should I take? Let us say this one. No. Yeah, let us say this is the one triangle. And then the second triangle I will draw. How do I draw? So, A and let us say this point I take. C and yeah, let me change this color as well. So, if I change the color for this one, let us say this one or this one. Yeah. Okay. Now, now I have to draw the third one. Now, here is the constraint. So, I know this angle, this is BDA, some measure it will be. Similarly, I can measure this angle, but I have to make sure that the third triangle which I am going to make is going to be 189. So, let me keep it here so that I will tell you why I am doing this. So, that I can make it easily. Okay. So, these are the remote angles E and D so far. So, let us measure these first. So, let me measure BDA. So, 53.97. I do not want these. So, this is but alpha. Okay. And this one AEC. Okay. So, this is 70.02 and this is beta. I do not need these names. Okay, alpha and beta. So, what I am going to do is I am going to construct all 180 degree angle here. So, what I am going to do is I am going to construct a line passing through E, which is at an angle alpha. So, this one is at alpha. Let me show the label. Let me not show the value or let it be value also. No problem. So, let the value be also there. And let me show this or rather I do not need it right now. So, I am going to construct a line passing through E such that AE and that line has an angle 53.97. How I am going to do it is this. So, I am going to reflect or rotate about a point. So, let us say about E I am going to rotate this line AE by what angle? So, by which angle? So, by clockwise direction? Yes, clockwise direction by what angle? So, the angle is alpha. Here it is. So, if you see this angle is going to be 53 point. So, let me show it here. So, how much is this angle? Let me measure this angle. So, this angle is A dash E dash 53.97, exactly same angle I have drawn that is gamma. Why did I do that? Now, so let me switch it off. Now, if I have a line let us say A dash E dash like that. If I see this line. Now, this angle here this one which one? Angle between let me measure that as well. So, let me measure this and yeah 56 point this delta. This delta let me yeah. So, this angle is delta here or let me show it here delta is 56.01. Now, I am going to find a point on you know a line passing through B and parallel to this line if I draw then I will get a point where okay. So, what I am going to do is I need a no not passing through B I need a yes I need a line parallel here. Why? Because see if I draw these parallel lines which one? So, let us say parallel line through C. Okay. Fair enough. Now, yes. So, this angle is if I somehow find out okay. So, what I am going to do is I am going to do this. So, I am going to draw a line parallel to B. Yeah, this is but will that help? No. This angle is alpha and I need a point here somewhere such that okay. See if I get a point here, oh no, but I am going to yes, yes, yes. I am sorry. So, what otherwise I could have done I do not need to do all of this. See what I need to do is once I get this 56.01 that is delta I need to extend not this line. So, this line is not needed. I need to actually extend this line which one? E dash yeah, perfect. Now, if I draw a line passing through B and parallel to this line then this given angle will be same. So, what I am going to see is so it says this. So, let me draw a line passing through B and parallel to this line. Yeah, perfect. Now, this angle clearly this one is my desired delta right. So, if I measure let me measure this angle. So, see 56.01. So, this angle is 56.01 and this angle anyways is beta and this is alpha. So, alpha plus beta plus gamma comes out to be 180 degrees. I hope this is understood. So, this is alpha here let me name it. Let me show the name. So, what I am showing is show the label. So, gamma is same as alpha my dear friends right 53.97 C this is gamma and this angle is beta 70.02. So, if you add all of them alpha plus beta plus gamma is 180 degrees right. So, this is remote angle beta this is alpha which is 53.97 and this one is my delta which is equal to here eta right 56.01. So, I hope you understood and we did construct this part where the remote angle sum is 180 degrees. Yep. So, these are the three triangles. Now, I need don't need these lines. So, let me just hide them. So, hide it I also don't need this. So, hide as well hide hide it and I don't need these all either. So, let me just hide all of them I don't need this I don't need this part as well this part as well right. So, let me just a minute this is yeah. So, e dash is also not needed. So, let me hide it. Okay. So, this is the thing and if I just do what I I know the point f. So, basically I need to draw this bf and fc. Okay. And I just need to or rather what I am doing to do is I am going to draw the polygon. Okay. So, polygon is bfc right and let me change the color of this as well. So, color. Okay. So, I will check. Yeah, let it be this color this color also has used to change it. So, let me change this to something like that. Oh, sorry, I have to change this one now. Sorry for the bad choice of colors if at all you think like that. We need to differentiate. So, hence I am doing that. Okay. Yeah, that will be too feeble. So, let me speed this and now oh wait. So, setting once again done. And this one done. Yeah, looks good. And this is let me just take this away and I don't need these either. So, let me so that my diagram is now needed. Okay. So, I constructed something which is now according to whatever you want it right. So, this is but this looks like it is a straight line but it will be a straight line. Yes, it will be a straight line anyways. So, can I randomize it more? How do I how do I randomize it more? So, let me say so this looks like ecf looks like a straight line. So, maybe it is appearing to be a special case. So, see I am going to do this I am going to draw a circle B passing through Bf C. Okay. And then yeah. So, I am now going to take any point here on this point. Let us say and call this as f. Okay. So, the angle will remain the same. So, I am utilizing the concept that angle in the same segment will be same. Okay. So, now let me draw a polygon from here Bf C. Now it looks perfect. Perfect. Right. And now I am going to yeah. So, this is the required third correct perfect. And then I don't need this circle either. So, let me just yeah. And let me switch it off as well. Okay. So, this is the construction guys. So, after all of this lot of work and we got what happens if I see if I change it. So, it doesn't matter. So, the angle is going to be the same. Yeah. This is all together will give will be same. It won't change it. So, I can yeah. This is the thing. Now let me see whether whatever is being said here then the circum circles of the three triangles meet at a point. Let me do and try that. So, let me draw the circum circles of the three. So, let's say the first one is this one. Okay. This is the first circle. Second one is this, this and this. And the third one is this, this and hey indeed it is meeting. Yes. See, this is the point of intersection where all the three are meeting. So, you can see that G. Right. So, the three circles semi-circles are indeed meeting at the point. And does it impact or is the position of the E point E guiding this? No. See, anywhere you take G point is going to be the point G here. Right. See, it's there. Always the three semicircles are, sorry, circum circles are intersecting. That's really cool. That's good. Yep. So, wherever you take these points irrespective of that. Yeah. This is beautiful. So, yes. Correct. Very good. Now, how do we prove it? How do we prove it? So, prove is not that tough. So, you can see. Let me prove it. So, how do I prove it? So, I am saying. See, out of the three, out of the three circum circles, two will definitely intersect. So, let's say AENC. So, this is the circle. This one. A, AEC. This circle. This is circle number one. So, let me write that as circle number one. This is circle number one. Right. This is circle number one. This is circle number two. Let's say. Circle number two. Now, definitely the two circles are going to meet at point C. One point is common point C. And another point is let's say G. Okay. So, let us join few lines over here. What all lines? So, let me join which one? So, let me make some cyclic quadrangles. So, A, G. See. Right. This one. And this one. This one. Yeah. This is good enough. Now, look carefully. What I am saying? I am saying is angle. So, the two, circle number one and two will definitely pass through C. There is absolutely no issues in that. Why? Because C happens to be one of the vertices of the triangle of which the circum circles are drawn. So, C will be the first common point. And let's say they intersect at point G. And at this point, I am not claiming that the third circum circle is also passing through G. So, hence what I am doing is let me do one thing here, which I can definitely do. And let's switch it off so that it doesn't create confusion. So, I am saying these two semicircles, circum circles are meeting at point G. Okay. So, what do I now know? I know that angle CGB, C, rather let me, yeah, CGB, angle CGB, this one, is equal to 180 degrees minus F, no problems in that. 180 degrees minus F. Why? Cyclic, quadrangle are opposite angles are opposite angles of cyclic, quadrangle, quadrangle are supplementary. Supplementary. Is it it? Okay, fair enough. Also, angle AGC, AGC, where is that angle? This one is also equal to 180 degrees minus angle E. Without doubt, same reason. Now let us try to find out angle AGB. Don't you think angle AGB, where is that angle? This one, AGB is equal to 360 degrees minus angle CGB minus angle AGC. Right? Angle at a point is 360 degrees. So, hence this is 360 degrees minus angle CGB. If you take CGB, which will be nothing but 180 degrees and plus angle F. And then this one will be 180 degrees plus angle E from these two. Okay. So, 180 degrees, 360, 360 goes. So, this is all cancelled. So, this is basically angle F plus angle E, angle F plus angle E. But there is an important information given in this question. And that is angle F plus angle E plus angle D, remote angles are put together is 180 degrees C is over here. That means angle F plus angle E is 180 degrees minus angle D. So, from these two relations, this is 1 and this is 2. You can very easily say angle AGB. AGB is equal to 180 degrees minus angle D. That means it is showing that AGB, this point G and D are, you know, angle AGB is 180 minus D. That means you can write AGB plus angle D is 180 degrees. What does it indicate? It indicates that the opposite angle of a quadrangle, these are opposite angles of two quadrangle. So, D and this angle G is opposite angle of a quadrangle and the sum is 180 degrees. Therefore, we can say that ADBG is a cyclic quadrangle and that proves the theorem. Why? The moment is a cyclic quadrangle. That means the circumcircle of ADG or ADB, ADB will pass through G. Isn't it? That is what, you know, because ADBG lies on a circle. So, hence there is, there can be only one circle passing through any three points. So, hence one circle which is passing through ADB will be the circumcircle of ADB only. There cannot be two circumcircles. So, circumcircle of ADB also passes through G. That is what we need to achieve and prove. And hence we can show this part, this circle activated. See, indeed this circle is the circumcircle of ADB and hence it passes through G as well. So, hence we could prove this theorem which says that if there are three triangles made arbitrarily on a, on the sides externally, on the, or constructed externally on the sides of an arbitrary triangle such that the sum of the remote angles again, what was the meaning of remote angle? ED and F is 180 degrees. Then the circumcircle of the three triangles meet at a point and that is this point G here, right? So, first we constructed this and then we prove this in the same session, right? I hope you understood this. So, if you did not understand I would request go back to the video once again and watch it once more and I am very sure in the second round you will be able to understand the entire theorem fully. Thanks for your patience for this long session and we will meet with another theorem in the next session. Thanks a lot. See you.