 Hello friends. So in the last session We had concluded that there are nine points nine points Within the triangle or on the triangle Through which a circle passes and that circle was called nine point circle We saw that in the previous session and in this session we are going to find out the relationship between the radius of this in this nine point circle and the circum center of or circum radius of the given triangle ABC and We have to prove that The radius of the nine point circle is half the circum radius of the triangle ABC Let's try and prove this in this session in Doing so we will also end up proving that the Center of this nine point circle actually lies on the Euler line of the triangle and what is Euler line guys the line joining the ortho center and the circum center of a triangle is called the Euler line and We will be seeing that the Center of the nine point circle also lies on the same Euler line Okay, so first thing to be you know Prove to you over here is if you see triangle ABC, let's consider triangle ABC. So We have triangle ABC triangle ABC and triangle D E F. Okay, let's consider these two triangles D E F Now if you see these two are similar triangles, we have proved it all you know already because this is a medial triangle if you see D E F is a medial triangle and We know that D F is equal to half of a C D F is equal to half AC midpoint theorem midpoint theorem and similarly D E is equal to half of AB half of AB and E F is half of DC Right, so by SSS criteria by SSS Criteria we can prove that triangle ABC is Similar to triangle D E F Right this would this proof will be needed in you know our larger objective over here So ABC is triangle ABC is similar to D E F Now if you look closely triangle D E F D E F is actually congruent to Triangle G I J Right, so G is opposite to D. I Is opposite to E and J is opposite to opposite to F right and why is this? Why is this because we know that in the previous session we proved that D F J G Is a rectangle? angle hence Hence, you know that F D Will be equal to G J Similarly, there are two more rectangles. You can check the previous session that You know, there are three rectangles in this figure one then two and third one was this Right, so three rectangles and bases that you can prove Similarly, you can prove that F E right, so similarly you can prove that F E F E is equal to I J I J and The third side what is left G I yeah, so G I G I is equal to ED so hence by SSS congruency criteria This holds D E F is congruent to G I J correct Now, let me just take away these highlights so that the figure is clear. Yeah Okay now so once Keep this in mind that these two triangles are congruent this one and And This one Right, please keep this in your mind now let's Let's consider Now let me do it here now if you see in it was given that BM BM I was given that BM is perpendicular to AC It's an altitude BM was altitude Similarly O E O E is perpendicular to AC why because O E was the perpendicular bisector of AC by the way if I forgot to mention this in the in this session I had mentioned in the previous one. Oh is the circum center guys So I'm just writing here. Oh is the circum center. Oh is the circum center Circum center of ABC Okay. Oh is a circum center of ABC sir. What is circum center? Circum center is point of intersection of all perpendicular bisectors So BM is clearly BM is perpendicular to AC and O E is perpendicular to a AC is a hence BM is perpendicular to we I'm sorry parallel to O E hence we can say that H I K is Equal to is equal to G that is equal to angle O E K alternate interior Angles correct now mmm If you see H happens to be H is the H what is H H is the ortho-center ortho-center of triangle ABC as well as as well as triangle a IGJ or IGJ why because if you see these are all 90 degrees This is 90. This is 90 This is 90. I'm sorry not that one So, yeah, this one and this one all are 90 degrees Why because G J was parallel to AC and this is 90 this is also 90 and This is also 90. So hence H happens to be ortho-center of both ABC as well as IGJ Why because if you see ortho-center is nothing but point of concurrence of our altitudes So H happens to be the point of concurrence of all altitudes of not only ABC, but also IGJ So H is the ortho-center of ABC and IGJ. Similarly O is the O is the ortho- center of Triangle E DEF O is the ortho-center of triangle DEF Why because O if you see This is perpendicular bisector. So this is 90 and G J was parallel to AC. So this is 90 as well and Similarly, this is 90 and I J is parallel to BC. So hence, this is also 90 Okay, so hence clearly. Yeah, O is the ortho-center of DEF now since triangle DEF is congruent to triangle What was that the G I J G I J therefore The distance between one of the vertices vertices and the ortho-center in respective triangles will also be same. So hence HI or in this case, it is OE OE OE is the Distance between vertex E and the ortho-center will be equal to in this case vertex H, sorry Ortho-center H and the corresponding side I Right. So OE is equal to HI. Right. Why because since these two triangles are Congruent, so hence the distance between the respective ortho-center and the vertices corresponding vertices will also be equal So OE is equal to HI Okay, OE is equal to HI. Now consider triangle IHK IHK and Triangle EOK EOK So what do we know in these two triangles? I know that HIK is equal to OEK you just proved above here Alternate interior angles then Clearly IK angle IKH is equal to angle OK IKH is equal to EKO, sorry EKO Vertically opposite angle Right EKO and we just proved that IH is equal to OE. Just proved here Correct therefore by AAS congruence criteria Triangle IHK is congruent to triangle EOK IHK is congruent to EOK, correct Right that means Therefore I can say IHK is equal to OK Right, so that means IHK is equal to OK and and IK is equal to EK CPCT both I CP CT right so that means what does it prove it proves first of all that K is the midpoint of midpoint of HK HO sorry And what was HO? HO is nothing but H is the orthocenter and O is the circumcenter of triangle ABC if you notice carefully so K is the midpoint of HO and K is the midpoint of IE that means K is the midpoint of IE and what was IE if you remember in the previous session IE was the dia of the 9-point circle right so hence K is the Center of K is the center of 9 Point circle right K is the center of 9-point circle and you can see the 9-point circle is the circum circle of triangle which is I'm writing which is the circum Circle of Triangle DEF as well right K is the center of 9-point circle which is the circum circle of DEF right therefore ratio of Circum radius circum radius of Triangle DEF Divide and or other let me write it like this circum radius of triangle DEF divided by circum radius of Triangle ABC ABC will be equal to nothing but by the law of similarity DE by DE what is the DE? DE by AB DE by AB is it it? Right, so two similar triangles since We have two similar triangles so hence it will be DE by AB So if there are two similar triangles all they are corresponding sides altitudes medians Circum radii all are proportional and hence it is equal to 1 by 2 Right, right, so hence we can say we can conclude circum radius or rather radius of Radius of 9-point Circle is equal to 1 by 2 times r r is Circum radius of Triangle ABC Okay, hence proved So I hope you understood the theorem. So I would suggest you just You know go through it once again You will get the entire method of this proof