 Hello and welcome to another session on gems of geometry. So in the previous session, we had discussed about various types of quadrangles. If you remember, there were three types of quadrangles which we discussed. And I told you that we are going to use these concepts or definitions and notations which we discussed in the previous session. Those were, you know, convex quadrangles. Then there was a reentrant, which is also called as concave quadrangles. And then there is something called crossed quadrangle. So in this particular session, we are going to discuss a very famous theorem which is called Wehringman's theorem. And you would have studied this in your 9th grade or 10th grade syllabus in congruent triangles as well. But you might not be aware of the name and the history of this theorem. So this theorem dates back to 1710, 1720s. So, you know, it was approved by Pierre Wehringman and he was a French mathematician. And this was published after his death, that is posthumously. So around 17, I think the year was 1731 when it was published. So if you get a chance, do a bit of research about Pierre Wehringman. He has contributed a lot in mechanics as well and calculus and Leibniz was his contemporary and Newton again. So Leibniz, Newton, Hooke and all these people were contemporaries. So he was, as I told you, a French mathematician. And he was the person who came up with this very simple but very elegant proof. Now, the Wehringman's theorem suggests that if you have a quadrilateral or quadrangle, as we had discussed, that we will be using the term quadrangle instead of quadrilateral. So if you have a quadrangle, A, B, C, D, I have shown you, this is a convex quadrangle, you can see that. And there is a concave or a re-interned, you know, quadrangle, I, J, K, L, right? Now, what the theorem suggests is if you take the midpoints of all the sides of the quadrangle and join them, you will get a parallelogram, point number one. And second is the area of the parallelogram so formed, that means in this case, E, F, G, H, and in this case, M, N, O, P, right? That, these two parallelograms are going to be, areas of these two parallelograms is going to be half of the original quadrangle, okay? So that means area of E, F, G, H is half of A, B, C, D. So area of M, N, O, P is half of I, J, K, L, okay? Both of them are parallelogram. The midpoints of the sides joined together, you will get a polygon which is a parallelogram in whether it is a concave or a convex parallelogram. We will also see in the next session, what happens when there is a crossed quadrangle. You remember crossed quadrangle was A, B, C, D, and then back to A. So this was kind of a, you know, twisted quadrangle which we discussed last time, is it? So what we are going to do is this, in this session, we are going to prove for convex and we'll see if time permits will do for concave as well. And then we'll also take a crossed quadrangle. But in crossed quadrangle, we will come back to negative areas, right? Where if you see the, you will definitely get a parallelogram over there as well. But the area of the parallelogram will be the difference of the two triangles. Or, you know, we will elaborate more when we discuss that because that will be a better way to do this. Okay. Now, so let's take up the quadrangle which is here, A, B, C, D. What I'm going to do is, first of all, show you that, yes indeed, it is a parallelogram all the time. So hence, if I move this point D, you can check. Let me just take this point D and move it around. So what I'm going to do is I'm going to move this point D around. So see, if I move this point D around, everywhere you can figure out that, you know, this is a parallelogram. Though by look, you cannot say that, but you can always see that they are parallel and equal. And one way to do it will be to measure the angle. So I'll show you one configuration. So let's say I measure this angle, 118.75, just keep a track of GFE now. So you can see both of them are, you know, equal. So that is one indication that they are actually a parallelogram. So let me take away this. And so hence, H is the midpoint of AD. E is the midpoint of AB. F is the midpoint of BC. And G is the midpoint of CD. And we have drawn a parallelogram. So how do we prove, basically I have joined the four vertices, but how do we prove that they are indeed a parallelogram? So here is where we will be using something called the midpoint theorem. If you remember, we had discussed some time ago. So what is a midpoint theorem? If you remember, there was a triangle. Okay. If there is a triangle and if you join the midpoints of the triangle, so the midpoints of the triangle are joined, two sides are joined, then this side, this particular, you know, segment is parallel to the third side. We know that. And also this is half of the base, right? So this, if this is X and this is Y, clearly X is Y by 2. So this was from midpoint theorem. So if you don't remember it, I would suggest you go to section one in the videos, in the description you will see the midpoint theorem and its explanation. Okay. So this is what we are going to use it. Okay. Now, so let's take the quadrilateral ABCD. Okay. So in this ABCD quadrilateral, E and H are the midpoints. So clearly EH is going to be parallel to what? BD. Simple. BD. BD is the diagonal which I have joined. Also, we know that EH is equal to half of BD because in this case, BD is the base angle. Now, if you look at this triangle, which one? BCD. Then what can I say about this? So FG. FG is parallel to, again, BD. And also, FG is equal to half BD. Now, we are getting a quadrangle where EFGH is the quadrangle and the opposite sides are equal. So if you see, EF is not only parallel to FGY because they are both parallel to BD. As well as EF is equal to FG and this is equal to half of BD. Both of them are half of BD. EFGH is a parallelogram. So I don't need to work hard on that. So EFGH is a parallelogram. No problem. I am writing in short time. So EFGH is a parallelogram. Then what theorem did we use? This was all based on midpoint theorem, or triangle midpoint theorem. Just have a look on midpoint theorem and it will become very easy for you. Now the second part is, the area of EFGH is equal to half of ABCD. How do we do that? So if you can see, for that, we will be needing one more theorem and that theorem suggests that the area of this top triangle here, if I just mark these lines over here. So this triangle EFEH, will be one fourth of triangle area ABD. How is that? So how is that possible? And you can use the similarity of triangles to prove that and we know that if you look at this diagram, angle A is common and this angle E is equal to angle B because of corresponding angles. So hence, can we not say that triangle AEH is similar to triangle ABD. Therefore, we can always say, we know a theorem which says that area of two similar triangles, AEH in this case. So let me write it little clearly. So what I am saying is, please pay attention is this, that area of triangle AEH, upon area of triangle, which one, ABD, ABD will be equal to the ratio of the, the square of the ratios of the sides, corresponding sides, right? So EH upon BD, whole square. This theorem we have studied already. Now EH by BD clearly is half, so this will be one by four. Okay, so we now know area of AEH is one fourth of ABD. So hence, first relation is area of triangle AEH is equal to half, sorry, one by four, of triangle ABD, triangle ABD, without any doubt. Similarly, you can all say that area of triangle, which one, EBF, triangle EBF is equal to one by four, yeah, one by four of triangle, which one, ABC, right? Triangle A, B and C. This is the second thing and third area of this triangle, this one here, this one, is how much, FCG, FCG, okay? So even if you don't mention the delta, it's okay. So one by four triangle, which one, BCD, if you check BCD, by the same logic and the fourth one I'm writing over here, is triangle area of triangle, which one, this one now, HBG, H, no, we'll have to write, anticlockwise, positive, if you're taking HGD, HGD is equal to one by four and triangle A, CD, right? Triangle A, C, B. I hope, this is clear, this is all area, so if I have missed writing AR, don't worry. Okay, so is that fine? So these are the four relationships I will get, AEH is one by four of ABD, EBF is one by four of ABC, FCG, CG is one by four of BCD and HGD, HGD is one by four of ACD, fantastic. Now, if you look at the parallelogram area, what is parallelogram area by the way, so area of, I'm simply writing EFGH, this is what we are interested in finding out, is how much? This is equal to, nothing but the full quadrilateral, that is ABCD and minus all the four triangles which we calculated over here, this one, this one, this one, not this one, this one, this one, this one, this one, and this one, right? If you see, ABCD minus AEH, EBF, FCG and HGD, if I take away all these triangles, then I will get the area of the parallelogram over there, is it right? The area of ABCD minus, let me write all of that, so one by four is common, so I can write one by four common and within brackets, what can I mention? Triangle ABD, triangle ABD plus triangle ABC, ABC plus triangle BCD, BCD, plus triangle ACD, ACD okay, so these are the two or four triangles here, now ABD, if you see, ABB plus BCD, right? So check, ABB plus BCD, this will give you another quadrilateral ABCD, is it it? So can I not write this same as area of ABCD minus one by four is outside the bracket and I will get one area of ABCD plus, if you check the other one, that is ABC this one, ABC and ACD, right? So these two combine and these two combine are individually giving you one quadrilateral each, you can check that, so again I will write area of ABCD okay, and close the bracket okay, so what is this? twice of ABCD and by four, that means this is half of ABCD so if you solve this, you will get half of area of ABCD and hence proved, that is what we wanted to prove that area of this parallelogram EFGH is equal to half of area of the quadrilateral triangle, right? Area of parallelogram EFGH is half quadrilateral ABCD, now this is for convex quadrangle but it is equally true for concave quadrangle, so what you can do is you can try this as an exercise and you will be able to get that so the process will not be much of a difference from what we have done here, so area of this parallelogram or MNOP will be equal to half of the area of IJKL right, so this is also you can establish, so try this as an exercise and let us know if you face any issues proving that, so hence in this session we could prove Warrington's term for a quadrangle which is convex right, so convex quadrangle we prove this term, we prove that if you join the midpoints of sides of the quadrangle you will get a parallelogram and the area of that parallelogram is going to be half of that of the quadrangle so in the next session what we are going to do is we are going to prove the same thing for a crossed quadrangle, so where you had seen that there is a twist given to the two of the sides and you get a crossed quadrangle and we will try and establish the same proof but the meaning of area of the quadrilateral over there would be slightly different which we will be discussing in the next session, I hope you like this one so see you again in the next session with the second part of the proof till that time, bye