 Hello friends, welcome again to another session on gems of geometry today We are going to do a little bit of interesting stuff. So so far we have been doing lots of theorems and proving the theorems. So today I thought probably we will be first drawing some figures and try to understand Geometry through you know drawing and construction. So today in this session, I'm going to discuss few topics like what is a medial triangle and then Also will be discussing something called Euler's line. So these are two, you know important vital concepts in a triangle So let's try and understand both them. So first of all, I am going to explain What a medial triangle is so first for that I need a triangle. So let me draw a Triangle, okay, so drawing a triangle now. So let's say the first point is this one and let's say this is a random point here and Let's say this is the Third point and fine point like that. So you can see a bc and three sides. It's a triangle Now we are going to draw the medial triangle. What is medial triangle guys medial triangle is nothing But if you join the midpoints of all the three sides You will get a medial triangle. So let's try to first find out the midpoint So I'm going to find out the midpoint of each one of them. So this one here is the midpoint Then again, I'm going to find the midpoint here and for here. This one is the next point Okay, so now if you join the You know, if you now draw a polygon from or let's say triangle from here to here to here so def is a medial triangle, okay m e d i a l medial Why medial because this is nothing but a triangle formed by joining the Feet of the medians of a triangle. So, you know that a median divides the sides of a triangle into equal parts or in you know, the median intersects the opposite side on At the midpoint of the side. So hence if you join all the three midpoints, you'll get the Medial triangle. Okay, so this is medial triangle. Now. Let's understand what Euler's line is okay. So for that, I just need to remove these medial track, you know, remove this medial triangle and then again Do some more construction. So I have removed the medial triangle now Let's now try to find out the Centroid what is the centroid guys centroid is nothing but a point of intersection of Medians of the triangle right so hence if I So this is so now let me Just Highlight this point. So if you see G is a centroid, right? G is a centroid G is usually the letter used for centroid as well. Okay, so centroid also is like center of gravity of The triangle if triangle is a you know, it's in a planar structure uniformly distributed mass is there Then G is the centroid. That means if you try to balance the triangle Such that the face of the triangle is parallel to the ground then you can balance it at point G Okay, so this is centroid now. We are going to draw the Orthocenter what is an orthocenter orthocenter is The line or sorry point of intersection of the altitude. So let's say this is the line So, let's say this point Let me just draw this point here. So BH is the you know Perpendicular or altitude so I will not use this letter H because it is useful for me later on So let's say this line is the line a line to be Altitude right another So hence, how do I find out orthocenter? So if I find if I drop a perpendicular from a on to BC right so if you see now I can highlight this point this point as you can see H is the orthocenter of The triangle I'm not drawing the third You know the altitude or the median in this case because you know that Or we already know that all the three medians are concurrent all the three altitudes are concurrent Similarly, all you know three perpendicular bisectors. Now. Now, let's draw the Circumcenter, so I'm now trying to draw the circumcenter of this triangle. What is circumcenter guys? So it's nothing but perpendicular bisector. So let's draw the perpendicular bisector of I'm sorry, wait a minute. So let's draw the perpendicular bisector Oh Sorry, Justin. So I'm now trying to draw a perpendicular bisector. So Here is the okay perpendicular bisector. So This one so this is the perpendicular bisector this line Right and then perpendicular bisector of let's say BC will be This one. So the centroid clearly is this point. I So we got three points HG. I'm sorry. Did I do something wrong? Yeah, I did something wrong because this is not Yeah, so this one is this one here. This one is I right the intersection of perpendicular bisector is called the centroid of the Triangle now Let's see an interesting thing. So now I'm trying to join H to G I have to choose it once again. So let me draw a Line Let me draw line from H on to G Okay, what did what do you notice guys notice this line this one? Yeah So Let me just yeah, this one this one highlighted one if you see closely this line is passing through H G and I that means Hg and gi Hg and I all are collinear. Isn't it amazing that in a triangle which I drew randomly the orthocenter the centroid and the Circumcenter are collinear now this line guys which joins the orthocenter the centroid and the The circumcenter is called the Euler's line. Okay, what is it called Euler's line? and if you look closely Orthocenter is farther away from the centroid than the circumcenter. Okay, this is one observation Which will be useful later and in the subsequent Sessions, we are going to also prove that Hg by gi. That means the ratio between the distance between the orthocenter and the centroid and the centroid and the circumcenter is 2 is to 1 that means this Hg is Twice of G1 How do we arrive at it? We will look or we will you know try to prove this in the subsequent Session so in this session, what did we understand we first saw what a medial triangle is medial triangle is nothing But the triangle joined or triangle formed by joining the midpoints of the sides of a triangle It's called the medial triangle and the Euler's line is nothing but the line Joining the centroid the orthocenter and the circumcenter and To our surprise these three points of a triangle which are very vital You know part of the triangle are collinear Interesting isn't it? So let's try and prove this particular thing that Hg and gi first of all We'll prove that this is valid for all the triangles that Hg and I are in this case Orthocenters centroid and the circumcenter are collinear and secondly will also prove that the ratio between The segments length that is the ratio of the segment Hg and gi is 2 is to 1 Okay, so let's meet in the next session