 Welcome friends to another session on gyms of geometry and in this session We are going to discuss one very important theorem called SIVA's theorem. Now this theorem was given by Giovanni SIVA Who published this theorem in 1678? Okay, so hence now let's like over four centuries now. I think yeah, so around three centuries plus We have been using this theorem. So what exactly is this theorem and before that, you know, you must understand what is Is meant by the word CVN. So what is a CVN guys? CVN is nothing but a line segment which joins a vertex of a triangle To any given point on the opposite side of the that particular vertex, okay? So for example here a x b y and Cz these three are CVN so hence what is a CVN any line which joins one of the vertex of a triangle Let's say a to any point on the opposite side. So hence, let's say if there is a point here and I Join this point. This will also be a CVN Okay, so this let's say M a m is also a CVN. So there can be infinitely many CVNs possible, isn't it? so now you understood what CVN is now SIVA's theorem says that The three CVNs a x b y and Cz one through each vertex of a triangle a b c are concurrent Then we get this relationship this one b x upon x c into C y upon y a Into a z by z b is one Interest interesting, isn't it? So hence basically these are the sides which are being, you know cut by the CVNs So the product of each in a cycle so b x upon Xc, so if you see this is one then this one by two Into three by four so the three by four are not numerals But the you know, so I'm just labeling these sides and then five by six if you multiply all of them One by two into three by four into five by six gives you one so that's what This particular theorem says now let's try to prove this theorem Okay, so how to go about the proof so hence in this concept in this particular theorem, we will be using the area Concept right so area of two triangles with the same base and between same parallel of its same altitude The two areas are equal in that case so hence if you see Let's take b x and xc. So if you see b x by xc If I take this ratio, what is b x by xc ratio? It's nothing, but I can write half Into b x Into let's say H. What is the H? B x into H. Okay, so H you can think of let's say there is a parallel line and The distance between these two is H. So half into half into base into height divided by can I not write this as half Into xc Into H Okay, let me write it once again So it is one by two into Let me put it in back it so that you know, you don't get confused it with multiplication sign xc Into H. Can I write that because? Half and H are common in both numerator and denominator. Now if you see closely this particular thing It's nothing, but area of triangle a triangle a b X right and the denominator is area of Triangle a xc a Cx let's say Okay, now let's say the three are the point of intersection is o a point of concurrence is o So this same thing can be written about o b x and o xc. So if you see Again, this it will be nothing but area of triangle o b x divided by area of triangle o Cx, why is this? Because if I write it here in brackets, what is o b x guys area of o b x will be let's say the height is Again some other h1 then o b x area is half into b x into h1 and If you see the denominator half into cx into h1, isn't it same? So h1 h1 goes half half goes so this ratio also is equal to bx upon xc. That's important Okay, so if you see from this one and Two so both are equal to bx upon xc. Let me write like that Bx upon xc isn't it bx upon xc both are equal now if two ratios are equal So they are correct difference of numerator and denominator So, you know component and dividend so hence I can write this by component or dividend or only dividend So bx by xc is equal to area of triangle a b x minus area of triangle o b x divided by area of triangle o c x Minus area. Sorry not this one will be a c x Minus area of triangle o c x clearly, right and why am I saying this because if a by b is equal to c by d Then if this will also be equal to a minus c divided by b minus d. You can check Yeah, let's take an example. So three by four is equal to let's say Six by eight Isn't it three by four is equal to six by eight So if you do the difference three minus six by four minus eight is equal to three upon four So all are same. This is just for explanation purposes. Nothing else So we get this relationship guys a bx upon xc is this now if you look closely to the figure in the figure area of a b x minus area of o b x will be nothing but dish shaded Region right so hence I can write now this as area of triangle a o b and in the denominator if you see this is nothing but AC x minus o c x is this particular area Right So hence you can write that as area of triangle triangle a o c or c o a Isn't it so what do we infer we we got bx upon xc is equal to area of triangle a o b divided by area of Triangle c o a and similarly if you repeat the process for the other three other two CVNs you will get C y upon y a Okay, C y upon y a will be equal to area of C y upon C y upon y a will be nothing but area of Bo a or Rather C y is first right. So hence. Yeah, it will be triangle B o c B o c divided by area of triangle a o b Triangle a o b right you can check you can the same process you will get the same result and similarly you will get a z by Z B if you see this a z by Z B will be a o c triangle triangle C o a This is area right area of C o a divided by area of C o b or B o c Right triangle B o c. So this is what you get now if you multiply all those three these three So hence, what will you get LHS multiply all the LHS together? So bx by xc into C y by y a Into a z by z B will be equal to Area of triangle a o b Divided by area of triangle C o a Into area of triangle B o c Divided by area of triangle a o b and this is multiplied by area of triangle C o a Divide by area of triangle B o c now all the areas are greater than zero in this case So hence we can cancel all of them B o c case goes C o a C o a goes Hence, this is one hence B Prove the result right so a x by xc into C y by y a into a z by Z B Right is equal to one. So this is what we needed to prove in fact converse is also true that means Converse if you just check converse, what will be the converse of this? So in a given triangle, I can word it out like that in a given triangle given triangle if if B x by xc is and into C y by y a into A z by Z B equals one equals one then then The three CVNs CVNs are Concurrent CVNs are concurrent this converse. This is the converse and you can prove it easily again So in this case, what will we do is let's say this is a triangle a b c and Okay Let's say two CVNs. Let's let's try with two CVNs first. So let's say B y and This one C z Okay, and let's say this point is X and it is given that what is given So B x by xc into C y by y a Into a z by Z B is equal to one. It's given isn't it? It's given so let's say at least two CVNs will definitely intersect if not all three So let's say this intersection is oh point point of intersection is oh, then what do we get? We know Then by CVN CVS CVS theorem. I can let's assume that there is a CVN Which is going on X But now this is very closely placed. So let me Remove this point and let's say this is the point X Okay, and by CVS theorem. Let's say we'll definitely get one CVN which passes through AO and let's say this is X dash so by CVS theorem by CVS theorem What do we get? We get this a z by Z B into B X dash by X dash C into C y by y a is equal to one But if you see what was given given was this so hence from let's say this is one and two from one and two from From one and two. What can we infer? We can infer B X by B X By X C is equal to B X dash by X dash C Now X and X dash are two points on the same segment BC And they are dividing the same segment into two equal ratios. Now. This is possible only when this is Possible only when X and X dash coincide right coincide that means X and X dash are the same point. So all the three hence hence hence all the three CVNs are Concurrent hence proved right so Converse is also True guys. So this theorem is widely applicable in various geometrical fields So you will see in olympic problems also this particular theorem is very very helpful so but then if you could see we didn't use any High-end mathematics to actually you know prove this result. So this result is Existing since 1678. So, you know, so there are more than three centuries now We have been using this theorem. So I hope you enjoyed this Theorem and you know in the worksheet attached with this there will be a few problem Problems to be you know solved using Siva's theorem. So best of luck for that. Thank you