 Welcome friends to another session on gems of geometry and as discussed in the previous session We are now going to prove that the nine point Circle exists right so in the previous session we saw what a nine point circle is and in this session We're going to prove it that it exists and all the nine points what all nine point nine point guys So the nine points are the three feet of the three Altitudes of any triangle. So let's say in this case M L and N the midpoints of the three sides that is D E and F and The midpoint of the segments from the three vertices to the ortho center namely G I and J right so what are the three? Nine points. I'm just highlighting or I'm writing here. So first is feet of the three altitude. So three altitude feet are M and And L These are the feet of feet of Altitudes Altitudes and clearly H is the ortho center H is the ortho center Correct in this case now What else the other? nice six points remaining are D E F D E F Are the vertices of the medial triangle? So D F are the midpoints? midpoints of AB DC and CA correct and I have I G and J these are the midpoints of These segments from the three vertices to the ortho center that means I G J are the midpoints of HB H B H A and H C I'm writing midpoint MP Okay, so these are the things and now we have to prove that all these points which all M and L D E F I G J all line the same circle, which is called the nine point circle Okay, the diagram looks a little intimidating But then we'll try to simplify it as much as possible now And we will be using the midpoint theorem a lot. So if you see Let's you know try to highlight one particular Triangle, so a rectangle rather and we'll see show you also that why it is a rectangle. So I Am highlighting this one Okay, now in this case now, let's understand What is happening here? So clearly F D is parallel to AC Is it why midpoint theorem midpoint Theorem right F and D are midpoints AC is the base similarly If you see G J G J is also parallel to AC I again midpoint theorem Midpoint theorem. Why because G is the midpoint of this G here is the midpoint of H A and J here is the midpoint of H C right so G J will be parallel to let's see. Okay Okay, now that means That's how similarly Similarly same logic D J is parallel to DJ is parallel to HB DJ is parallel to HB midpoint theorem and G F is parallel to G F is parallel to HB midpoint theorem therefore DJ is parallel to HB or rather G F DJ is parallel to G F. Hence we can say G F DJ is a parallelogram parallelogram right Opposite sides being parallel so parallelogram now if you see G F G F was parallel to HB and And HB if you see is perpendicular to Right HB or HM is perpendicular to AC Is it if you see HB this one this HB here. Let me just highlight it HB is HM and HM is perpendicular to AC is it and and AC is AC is POP and AC is parallel to G J We just proved Therefore, what do I what do we mean? What do we understand from here? So G F is perpendicular to G J Right therefore angle F G J is equal to 90 degrees and hence therefore G F DJ is a rectangle Any angle of a parallelogram is 90 that means the rectangle G F DJ is a rectangle that means what can I infer if fj consider the diagonal for for Diagonal fj okay for diagonal fj D and G are points Lying on a circle lying on a circle with Fj as diameter So if you consider fj to be the let's say this is fj So if you see this point and this point are 90 degrees So hence they will lie on a circle passing through all these four points such that fj is the diameter I hope you understood this Okay, similarly, let me just remove this similarly Now if I now let me write here so if you now see so fj is the raya and And D and G are the points on the circle right so four points these four points are definitely consigned it correct now Now yes, so if you now see let's Similarly, I can prove see me Lerly I can prove f i J and e F i J and e is a rectangle Rectangle F i parallel to a l Midpoint theorem e j parallel to a l Again midpoint theorem So hence e j is parallel to f i and similarly i j will be parallel to an e so Parallogram and again you can prove that there's a 90 degree. So f i j is a rectangle. So hence F j as a dia if f j is a dia is a dia then I And Lie on the circle Right, that means f j i e Are consigned click So what do we learn? So if f j if you consider f j to be the dia then we just proved that points What all D? I E and G these are all the same circle now the only two points left to be you know this thing is nn Point N and L if you see now, isn't it? How do we do that? Okay, so so If you see a rectangle rectangle F i j e f i j e then Then I e and F j R d dia both Are the dia of what? Circle with circle of the circle with circle which is passing through all these you know which Passes through passes through I f I f j I And e Okay, right. So now similarly if again we can again prove similarly You can see I I D Sorry, yeah, yeah ID and EG so ID and I D e g EG is a Rectangle same thing same by same logic you can see ID e G is a rectangle also guys if you see Angle F N J is 90 degree why because this is 90 degree because Cn is an altitude right so f n j is 90 degrees so hence you can see if F j is a Daya of a circle Circle then N will Lie over that over the circle Right, why are we saying all this because if you remember guys if there is a circle and There's a diameter. So any point on the Circle or Any angle subtended by the dia on the circle will be right angle. We have studied this in our basic geometry Right. So all these are 90 degrees. So wherever you see 90 degrees with one Daya then all of them will be on same circle, right? So similarly n is on the same circle and and similarly angle D L g D lg is equal to 90 degrees 90 degrees so so if If a Circle be drawn the circle be drawn with G L as dia Then D will be on it D will be on It right so we see that N N I'm sorry L will be on it L will be on it Right. So N L L are On this, you know, so if you see now Fj fj and gl fj and gl Where the I'm sorry Let is some some mistake here. So if a circle be drawn with gd my bad Gd as dia, right? Gd as dia, please do a correction If a circle be drawn with gd as dia, then L will be on it, correct? So now what are all these circles so gd was the diagonal of a circle or rectangle fdjg When which fj is also the dia so hence dg is also the dia and there's another dia which is ie So all these are dia, right of the same circle Which are having points f i d j e g on them only points left to be proven to be on circle are L N and M now we just prove that L and N are also in the same circle. Why is N on the same circle? Because N Fnj right Fnj is 90 degree fnj is 90 degree. So if fj is a dia and will be on the circle similarly D lg is 90 degree, isn't it? Why because AL is the perpendicular so D lg is 90 degree So if Dg is the dia or gd is the dia then L will be on it, right? Now Dg and fj were the diagonals of same rectangle fdjg. So hence Fj and Dg are the dia of the same circle, right? So hence L and N will also now lie on the same circle, right? Now the only thing remaining is M's but by similar logic you can prove that similarly similarly M will lie on a circle on a circle with With ie as the dia ie as the dia. Now ie and Fj were the Diagonals of the same rectangle. So ie and fj happens to be the dia of the same circle, right? So hence hence hence M is Concyclic with L and N Hence if we now see all the points which all D e f D e f then L M N and and I g j or g j i Right all are Concyclic. So hence we could prove that we can we could prove that all the six points or nine points are Concyclic, okay? So again, just a quick recap There were three rectangles we proved because of midpoint theorem. So I will highlight them once again So one is already highlighted another one is this rectangle we proved and The other one was this one Right, so these are the three rectangles which we proved. Okay, so all these points will be on the same Circle we proved that and after that we've also proved that N L M are also Lying on the same circle with ie or Gd or fj as the dia correct and hence all the three all the nine points Sorry, are constantly in the next session. We'll also prove that There is this half of our