 welcome friends so we have been seeing in the previous few sessions the properties of a triangle and its medial triangle and you know in this session we are going to prove that the orthocenter centroid and circumcenter of any triangle are collinear first of all and then the centroid divides the distance from the orthocenter to the circumcenter in the ratio 2 is to 1 okay so the diagram looks a little intimidating but don't worry we'll try to explain each and every bit of it so let's start so let's first understand what this theorem is talking about so it says orthocenter what is orthocenter guys so you know orthocenter is nothing but point of concurrence of all altitudes of a triangle so clearly if you see in this diagram I have not shown all the three altitudes because we know that if two altitudes are intersecting the third altitude will pass through that point of intersection only right so hence for the you know so just to declutter the diagram I have only shown two altitudes here so namely BK is one altitude and AL and AL are the altitudes here are the altitudes so I've just shown these two and clearly H becomes is the or H is the orthocenter right so this is first understood I believe now what's a centroid centroid is nothing but where all the three medians meet so again by or for the sake of clarity or decluttering the diagram I have shown only two medians one is AD so AD and BE, BE are medians and we know that all the medians are concurrent all the three medians of a triangular concurrent and in this case G is the centroid ok so we know this and then third is the circumcenter what is circumcenter guys point of intersection of all the perpendicular perpendicular bisector ok so perpendicular bisector so let's understand now so in this case if you see EJ EJ and FI are perpendicular bisectors and we know that the point of intersection of all the perpendicular bisectors is called circumcenters for I is circumcenter circumcenter though usually we denote circumcenter by letter O but never mind so in this diagram we're going to call it I okay now we have to prove that HGI so we have to prove that HGI are collinear first of all okay and secondly second is HG upon GI is 2 upon 1 this is what your theorem is demanding us to prove so let's begin guys so how to prove so let's first try and understand the you know some of the properties which we know so by midpoint theorem we know that yeah from midpoint theorem we know that FD FD is parallel to AC midpoint midpoint theorem what does midpoint theorem say it says that if you join the two midpoints of two sides of a triangle then it will be parallel to the third side so FD is parallel to AC okay similarly ED is parallel to AB okay now if JE or EJ is perpendicular to AB or AC then then EJ is also perpendicular to FD because FD and AC are parallel right so I have shown that here if you see I will just encircle the this is the 90 degree here 90 degree here right so these are 290 degrees okay and now what BK is perpendicular to AC right therefore what do we conclude BK and other perpendicular on AC that is EJ so BK is parallel to EJ two perpendicular are always parallel BK is parallel to EJ that is what so hence this will imply that angle H B H BG H BG will be equal to angle G E I G E I let me just clear this encircle stuff so that yeah GE I I hope it is visible to all of you okay so GE I this is what and what is the reason behind it so you can always mention this is alternate interior angles fair enough so right so this is what we know now we also know let me just write it here let the diagram be here so we know what we know that medians divide each other in the ratio medians divide each other in the ratio pushed one or other I should say centroid divides yeah so instead of writing this I should be writing let's say centroid the centroid centroid divides the median centroid divides the median in the ratio two is to one that means BG I can clearly say BG upon GE right BG upon GE is equal to two is to one five we know this now if you also look at this and let's try this you know triangle ABC is similar to triangle B E F isn't it why because if you see angle BAC BAC is equal to angle EDF angle EDF right why opposite angles of a parallelogram are equal correct so if you see this angle is definitely equal to this angle why because AEDF happens to be a parallelogram by midpoint theorem we already saw if you can see here these two are parallels right so that means this will clearly say that AF ED A F DE is a parallelogram so D is equal to angle A clear similarly you can always say with the same logic angle DEF will be equal to angle ABC therefore by AA similarity you'll say right triangle ABC is similar to triangle DEF now midpoint theorem also says that FD by same midpoint theorem by midpoint theorem we know FD by AB is equal to sorry FD by AC rather FD by AC is 2 by 1 isn't it and by you know we know that if two triangles are similar then the distance there are you know so if two triangles are similar then the ratio of their median their or their altitude their angle bisectors all are proportional to the sides what am I saying that if these two triangles are similar then their altitudes their centroid and their sorry their altitudes their angle bisectors and their medians all of them will be proportional right and hence by that logic we will conclude that if you see BK BK is the altitude of ABC and EJ is the altitude of DEF correct right so from there from this particular concept or theorem that all the corresponding parts of two similar triangles are always proportional hence we will say that BH BH by IE okay BH is the orthocenter and from H to the vertex B is BH and in the second triangle I is the orthocenter I happens to be the orthocenter if you see because these are right these are perpendicular for example if you see let me say this point is M okay so FM also will be perpendicular to ED why because ED is parallel to AB by midpoint theorem and hence propellant FM or Fi was perpendicular on AB so it will be perpendicular on ED as well right so hence if you see I is the orthocenter of DEF correct so hence if you take the ratio of the distance from the vertex B to its orthocenter H with the with the distance of E from the second orthocenter of the triangle DEF which is EI here so I am just trying to tell you is this that this particular part versus or this particular part these are the corresponding distances of orthocenter of the respective triangles from the vertex correct so by the logic that for two similar triangles all the sides all the altitudes all the median all the in you know in angle bisectors and all such corresponding parts will be proportional we will see BH by EI will also be equal to 2 is to 1 why because ABC is similar to DEF okay ABC is similar to DEF then now consider in triangles now HBG HBG and triangle IEG IEG okay HBG and IEG we can say angle HBG HBG is equal to angle IEG IEG which is we just proved here if you see this is the thing given we just proved then similarly angle what else and this the first one was alternate interior angle and the second one we can and not angles now anymore so we will be comparing this you know ratio of the sides so we now just learned that BH by EI is equal to 2 is to 1 is equal to 2 is equal to FD by FD by AC FD by AC or sorry BG by GE this one this one will use right so we will write BG by BG by GE so hence my dear friends we can say by by ASA similarity rather SAS my bad SAS similarity by SAS similarity triangle HBG is similar to triangle IEG right HBG is similar to IEG so hence what can what can we conclude now AE is BE is the median guys this one is the median right this one is the median and it's a straight line correct and since these two are these two triangles are now similar we can say that angle HGB will be equal to angle IGIGE right why corresponding parts of similar triangle right so that is true and we can we know that BE is a BE is a straight line is a straight line because it's a median therefore this particular condition is true only when HGI is a HGI is a straight line straight line that means therefore HGI are collinear right proved first part first part is proved now let's see the second part now since it follows directly from the same you know the similarity itself that HG by GI will be equal to 2 by 1 why because the other two sides are in the ratio of 2 2 is to 1 so in this case the third side HG and GI will also be in the ratio of 2 is to 1 hence proved hence hence hence hence proved right it's a very interesting proof guys you know I know some of you would be you know kind of puzzled when we did this but then I'm telling you you can just we'll put another video where which explains that corresponding parts of similar triangles are always proportional to its sides so but otherwise you know I think you would have understood the theorem just to reiterate the orthocenter centroid and circumcenter of any triangle are collinear and the centroid divides the distance from the orthocenter to the circumcenter in the ratio 2 is to 1 I hope you like this session