 Hello friends and welcome again to another session on Gems of Geometry. Today we are going to take up another very interesting theorem and the theorem's name is Mineloss's theorem. Now Mineloss is a philosopher or a mathematician. He was in Alexandria around 180 in around 180 and you know that Alexandria was part of Greek Empire that time right. So this person you know has tremendous contribution towards geometry and one of his contribution is Mineloss's theorem his own you know this this theorem. Now what does this theorem say? This theorem basically deals with if I have to say a co-linearity of points co-linearity of points. So what I have done here is I have first done the construction and first let us understand the theorem then we will verify it then we have to prove it. So what does the theorem says? The theorem says that if points x, y, z so if you can see this is point x here this is point y and this is point z on the sides b, c, c, a, b so you can see b, c and hence you have to extend if it is not lying on that line. So I have extended b, so x lies on b, c extended y line y lies on a, c and z lies on a, b. So if these are three points and if they are co-linear so I have mentioned that if they are co-linear then what will happen? So the ratios b, x upon c, x right. So this is b, x upon c, x first one. Then c, y upon a, y so this one again c, y upon a, y and a, z upon b, z. So you can see there is an order I am going from b to x to c then from c to y to a and then from a to z to b right. So in only one case I have to come back otherwise I do not need to come back but what I mean is I go to b from b to x I go like that and I have to come back towards c is not it? I have to turn my direction change my direction but otherwise if you see c to y, y to a and a to z and z to b I am not changing direction. Later on we will also see whenever change of direction is concerned we attach a minus sign to represent that you know there is one change in direction all the time or it will be in other case we will see when the line does not intersect the sides internally and all of the and the line intersects externally then all of them in all the cases we have reversal of direction. So why do not I just show it to you? So what I mean is let us say if I have this line half you know around it like that so if you can see it is it is going out right so let me just yeah so what I mean is look at this line now in all the three cases it is externally dividing right so this line is dividing the three extended sides of the triangle externally so you can see the point x is outside on extended v c point y is on extended c a and point z on extended a b. So in all the three cases here I have to change the direction what do I mean so let us say if I go from b to c or sorry x then I come to x to c then again from c to y I have to go and then come back to a and then again I have to go from a to z and come back to right so there are three times I am changing the direction so hence three negative sign so all the time you know you know I am changing the direction so you will be changing either three times or simply one time you cannot change direction only twice so meaning thereby this line is either going to cut two lines at time or cut none at all so that is the thing let me just change the position of this line to show you what I mean so therefore either only you know it is cutting extended lines now it is cutting the two lines or two segments of the triangle internally and one one here externally is it and there is no other possibility right no other possibility again here two internal one external and then again all three external right so three external one external or three external three external one external three external so there is no two external cutting right so this is the scenario now what I am going to do is I am going to verify this so that it is indeed one so you can see it has just changed its location so let me put it here now let me pin it okay now it is not going to change the position so if you if you see I am switching on the dimension and let me bring it back so all these are the dimension I have switched on the dimensions you can see the ratios are indeed coming out to be one right indeed coming out to be one so how do we prove it let me just put it here and let me put it here okay so wherever you want to change it the value is always one so that is mathematically calculated here now let us prove this for this let us declutter it so I am taking away all the dimensions let it be here and then I am doing some construction so what construction I am going to do I am going to drop perpendicular like that so you can see cf is perpendicular to the line and ag is perpendicular to the line and bz or bh is perpendicular to the line okay now don't get confused let me just take this point here and this point is z here okay now how do we prove this here and the proof will be pretty simple so now let us talk about its proving so how and what should be the approach for the proving guys so what are we going to do we are going to take help of similar triangles so let's discuss its proof in detail okay so you can see we have done some construction where cf is perpendicular to the line ag is perpendicular to the line and bh is also perpendicular to the line so let me mark the perpendicular okay now in triangle or rather I can say triangle xfc look at the triangle xfc and it is similar to triangle x h and b yes or no why because x is a common angle to both the triangles so xfc is one and x h and b is the other one so x is a common angle and there is there are two 90 degrees so here is one and the other one is here this one so both are 90 degrees so hence by a similarity criteria right it is there that these two triangles are similar if the triangles are similar then what can I say about the ratio of the sides we know and I would write if I write let's say xc or cx cx by bx what should I write cx by bx is equal to so this is cx this is bx whole so it will be equal to cf by cf by bh is it it cf by bh so these are this ratio will hold then similarly let's see triangle agy and yfc these two triangles okay so I am saying triangle agy is similar to triangle c fy why because again if you look at angle a angle angle y rather this angle y and this angle y is same vertically opposite and this is 190 degree here and another 90 degree here so again by a similarity you can say that these two triangles are similar so hence you can say what you can say ay by cy ay by cy is equal to ay by cy is equal to ag by cf right similarly isn't it so we got ay by cy is equal to ag by cf now third one triangle look at this triangle which one so let me shed it for you so this one this one and this small triangle here right so in this triangle if you see triangle let us try say bhg is similar to triangle a sorry bhz not g my bad so this is bhz here so z is hidden if you if you look at it this is this is z this is point z so bhz is similar to agz agz and hence we can say that bh by or we can say rather um az az by bz az by bz is equal to az by bz az by bz is equal to ag by bg bh sorry bh ag by bh right now these are the three ratios which we get and let's now say that this is nothing but if you reciprocate you will get bx by cx is equal to bh by cf no doubt about it and this is in this is also if you receive reciprocate you will get cy by ay is equal to cf by ag is it not so let us say this was relation number one this is two and this is three as well okay now what do you need to multiply one two and three together so lhs if I product them all I will get bx by cx into this that is cy by ay into az by bz so this gives me the left hand side of the desired result vx by cx cy by ay az by bz now let's multiply the rhs of all so this is bh by cf into cf by ag by now you would have realized that cf is getting cancelled let's see the others are getting cancelled or not so ag by bh is it so beautiful so this is cancel all are canceling each other out so what is left we are left with one right so hence we can see that if there are three points which are collinear and the line is cutting the triangle like that then the ratios in which the points divide the line whether internally or externally the product of all the three ratios will always be one so this is how we proved it actually the converse is also true and the converse can also be stated as if there are three points such that this ratio holds then the three lines are collinear right so hence millenosis theorem can be used to prove collinearity of points right and the proof for the converse will also be very easy so what you need to do is you consider x and you join x and y and let it be produced to some point here z dash okay and then by you know again by contradiction you can always prove that z and z dash will coincide so we will look at the proof in the next session but this is how millenosis theorem can be proved right so millenosis theorem as I told you can be proved or this is this is a theorem for related to the collinearity of points and it's and their association with the triangle okay so I hope you understood the theorem and its proof let's move on to the next proof thank you