 Hi, I'm Zor. Welcome to Unizor Education. I continue with problems. This is the third series of problems related to quadrangles. Actually triangles and quadrangles they're bunched together and obviously all these elements are used in all these problems. So this series contains nine problems and let me just go one by one. Prove that in an equilateral triangle some of distances of any internal point to all three sides is constant and equal to the altitude. Okay, this is equilateral triangle and we need to prove that the distance from a point to these three sides, some of these distances, is the same as an altitude. Now if you remember in one of the prior lectures we had a similar problem when you had a socialist triangle and any point on the base began two distances from two different sides and we have proven that some of these is equal to an altitude to a leg. Now I'll just use this right now. Now how can I use it? Well let's draw a parallel line through this. Now obviously since this is parallel to this these are two perpendiculars so these two segments congruent to each other which means that they have to prove that some of these two is equal to this piece. But this is exactly the same as this particular theorem except in this case it was an altitude towards the leg but in case of equilateral triangle all altitudes are the same. So basically this distance from this to this is the same as from this to this. All altitudes are the same. So that finished basically the proof. And it's kind of common that if you have solved one problem then you have another problem which you can reduce to the previously solved one. Well good, I mean that's the easiest way. It's not the how many little steps you are making towards a proof or solution or construction or whatever. What's important is whether you can or cannot do it. And in this case the easiest way to do is just to reduce the problem to previously solved one and basically forget about all the little steps which involved in solving this previous problem. But solved itself can be used and reused. Prove that parallelogram with congruent diagonals is a rectangle. Alright, so you have parallelogram with congruent diagonals. We have to prove that this is a rectangle. Alright, what can we do? Well for instance we will draw a line parallel to this diagonal here. Now since these are parallel and these are parallel, obviously EB and AC are parallel and equal to each other. Now since diagonals are congruent, then EB and BG are congruent. Now since this is equal to this and equal to this, it looks like BA is a median in isosceles triangle BGE. BGE is isosceles because this diagonal is equal to this diagonal and this in turn equals to this one. So BGE is a isosceles and BA is median because AG is equal to AE. Now in isosceles, in the isosceles triangle, median from the top to the base coincides with angle bisector and altitude. That's why this is 90 degree and this is 90 degree, which means we have a parallelogram with 90 degrees internal angle, which means all angles are 90 degrees because opposite are congruent to each other and neighboring angles are supplemental to each other and supplemental to 90 is 90 degrees as well. So all 90 degrees, so that makes all angles equal, right angles, and that's a rectangle, by definition by the way. Remember definition of the rectangle is a parallelogram with all internal angles equal to each other. So they are all equal to 90 degrees. Prove the parallelogram with perpendicular diagonals is a rhombus. So again we have a parallelogram and diagonals are perpendicular to each other. So these are right angles. Okay, they are perpendicular to each other, but now let's look at these right triangles. A, B, C, D, M, B, M, A, and A, M, D. Both are right triangles. The tragedy are congruent because in a parallelogram diagonals are intersecting in the middle. This one is common through these triangles and that's why triangles are congruent and that's why this particular hypotenuse is equal to this one. Now, and this in turn is congruent to B, C because it's a parallelogram and A, B is congruent to C, D because it's a parallelogram. So all four sides are congruent to each other and this is a definition of a rhombus. Prove that parallelogram with the diagonal being an angle bisector is a rhombus. Okay, very similar, but now instead of perpendicularity of the diagonals I have that these angles are congruent to each other because this is a bisector. Okay, that's good. Now, this angle is equal to this obviously because these are parallel and this is transversal. So these are alternate interior angles which makes a triangle A, B, C and the one with two angles at the base congruent to each other and we know that these triangles are isosceles. So that's what makes these two sides congruent to each other. A, B is congruent to B, C because A, B, C is isosceles triangle by two congruent angles at the base A, C. And obviously because it's a parallelogram, B, C is equal to A, D, A, B is equal to C, D, all sides are equal, that's the definition of a rhombus. Given a rhombus from a point of intersection of its diagonals with dropped perpendicular to all four sides, prove the points of intersection of these perpendicular forms a rectangle. Okay, since it's a rhombus, let me just draw it this way. Better, a more acute angle. Okay, rhombus. Now, we have two diagonals, one and two, and from the center we dropped perpendicular. One, two, one, two. So we have to prove that the end points of these perpendicular form a rectangle. All right, let's think. Well, since it's a rhombus, all four right triangles A and B, B and C, C and D and A and D, all four right triangles are congruent to each other. Now, they are right because rhombus has perpendicular diagonals, we know that property of the rhombuses, and we also know that they all share casualties and all hypogenesis are congruent to each other because it's a rhombus. So obviously triangles are congruent, which means all these altitudes in each of these triangles are congruent to each other. These are corresponding altitudes from the right angle to hypogenesis. So if they are all congruent to each other, that makes our figure a quadrangle with two diagonals crossing in the middle and congruent among themselves. Two diagonals crossing in the middle and congruent to each other. Now, and we have already proven many times that this is a rectangle, this the square triangle which has this property of the diagonals. So again, we reduce the problem to previous solved ones and basically shorten our way. Actually, the whole mathematics is built this way. You always try to prove something new based on whatever has already been proven and you don't go back to the beginning to the axioms to basically prove and prove again the same thing. You just use whatever you have already proved. Okay, prove the angle by sectors of a rectangle for a square. This is more interesting. So you have a rectangle and you have by sectors. Well, if I did it a little bit more precisely, then I would say that this thing E F G H form a square. Now, the point G is not necessarily on A D and F as well. I mean, they can be somewhere in the middle actually. So how can we prove that this is a square? Well, first of all, obviously, it's a parallelogram. Why? Because these angles are congruent because of the parallel lines and transversal. And these angles are also congruent for the same reason because these lines are parallel. And this is transversal. Okay, and they are all equal to 45 degrees because these are all angle by sectors. I don't even have to prove anything related to parallelism. I just say this is 45 and this is 45. So these are equal and these are alternate interior. So this is also 45. This is 45 degrees. Alright, so we have and this is now if we will consider A F D, a triangle A F D, you will see that this is a triangle with two angles equal to 45 degrees, which leaves this to be 90. And basically the same thing for this side also 90 degree. We just have to consider B G C triangle. This as well because B E A triangle has 45 degrees and 45 degrees. So this is 90 and this is vertical to 90 which is equal. So we have a rectangle with all that all interior angles 90 degree. That's exactly what makes it a rectangle. Okay, given a square A B C G, A B C G square points A prime is middle of C D, A prime, D prime, C prime and D prime. Now we have to prove that this is a square. So we have to prove that M N P Q is a square. Okay, let's think about it. Oh, not only a square, but also that each side of this square is equal to two-fifths of this seventh. Like in this case M Q is two-fifths of B B prime. Alright, so how can we prove that? Well, let's do it this way. First of all, we know the theorem that if you have an angle and you have two congruent segments on one side of the angle and we draw two parallel lines, then on another side we also have congruent segments. So we will use this theory. It was proven before in one of the lectures. Now how can we prove it? Well, since B C prime is equal to C prime A and these lines are parallel to each other, which probably needs to be proven actually. Yeah, let's start from the parallelism. That would be interesting. So why are these parallel? Well, probably we can say it based on the congruence of the triangles. A A prime D is congruent to D D prime C congruent to C C prime B and B D prime A by two casualties. Two casualties. One is equal to half the side of the square and another is the whole side of the square. And they're all right triangles. So two casualties enough to prove their congruence, which means that these angles are to each other. Okay, since these two angles are congruent to each other, that's why... Oh, okay. Before we do this, since we have proven that these triangles are congruent to each other, and using this theorem, here is why, these two angles, this plus this is equal to 90 degree. Why? Because this angle is equal to this. And these two in sum gives you 90 degrees. Because C D prime D is right triangle, and we have already proven that this angle is congruent to this one. Since these two are two acute angles in a big triangle, these two angles make up 90 degrees as well, which means this is a 90 degree angle. And in exactly the same fashion we proved that this is 90 degree, and this is 90 degree, and this is 90 degree, which means whatever we have in the center is rectangle, because all angles are right angles. They're all vertical to these guys. So, let me just start again. Let me repeat this particular watch. D G prime C is right triangle, because, you know, this is the right angle. Now, these angles are congruent to each other, because this right triangle is congruent to this right triangle. They have exactly the same pattern thing, which means that since these two angles of this triangle make 90 degrees, these two angles also make 90 degrees. And in a small triangle D prime C N, this remains to be 90 degrees. Okay, so all angles are 90 degrees, so this is a rectangle. Now question is, why is this particular rectangle is a square? Why sides are equal? Okay, okay. Now since these two angles make up 90 degrees, and these two angles, this and this, make 90 degrees. So these two angles C prime C D and A A prime D are congruent to each other, because being added the same angle here or here, they make up 90 degrees. So they are the same. And since they are the same, C C prime and A A prime are probable to each other because these are corresponding angles. And same thing everywhere else. So the lines are parallel, which means since these two segments are congruent to each other, these two should be congruent to each other because this is an angle and we cut it through with two parallel lines. So this one, side of the quadrangle in the middle, has exactly the same side as this one. But now, let's think about this one. These are exactly the same for obvious reasons, to the same reason, because these are equal to each other. And since these are also the same, these are also the same. Now, so these triangles D prime C N and B M C prime also are congruent to each other because they have an acute angle and a hypogenous the same. So this is also two and this is also two and this is two and this is two strikes. So these are all congruent segments. So I have basically proven that this is a rectangle and this is a rhombus because all sides are congruent to each other, which means it's a square. So the only thing which remains to be proven is that this length is two-fifths of the whole C C prime. Well, what I would like to prove that this particular piece M C prime is half of this. How can that be proven? Well, very easily. Let's draw a line here and consider this is C double prime. Now, B M C and C C double prime A are congruent because since these are parallel, the angles are congruent and hypogenesis are exactly the same because C prime is a midpoint of AB. That's why this is equal to this. But at the same time, since this is equal to this, since it's a parallelogram, this small C M Q C double prime, so what we have is we have this particular piece basically divided by two equal segments, each of them congruent to this one. So this one is half of this and in turn this piece and these pieces are all the same, right? They're all the same. So every short piece here, here, here and here, every short piece is half of the square and square has the side equal to this one. So we have one segment plus one segment, this plus this plus half segment this one. Whatever the segment is, you can use like x plus x plus x divided by two and what do we have here? Well, this is obviously five seconds. This one being two seconds of the segment, which is two-fifths. So this is two-fifths. So if you divide this by two, we have one, two, three, four, five pieces and two pieces are on this side. No, the drawing is a little messy. I hope you understand. Okay. Okay. Given square a b c g and points square a b c g against square and points a prime, b prime, c prime, c prime positioned inside a b corresponding with size of the segment. Okay. This is on the same distance as this on the same distance as this and on the same distance as this. So a a one is congruent to b b prime, I mean c c prime and g g prime. And now we connected, connected these points and we have to prove that this is a square. Well, let's think about this way. Since it was a square and we have already made these segments congruent to each other, that means these segments are also congruent to each other because each big segment represents the size of the square minus something which is the same on all four sides. Now that makes all our triangles, right triangles mind you because this is a square. So these are all right angles congruent to each other, which means that every side is congruent. So that's number one. Which means this is a rhombus. Now next, how can we prove that the angles are the same? But obviously, angles are the same because from 180 degree angle, we subtract one acute angle, a smaller acute angle and a bigger acute angle. One from this triangle and one from this triangle. And here we do exactly the same. From 180 degree we have to subtract one small acute angle and one bigger acute angle. And we always get the same difference. Same thing is here. We subtract a small acute angle and the bigger acute angle and since ultra angles are congruent to each other, we always have exactly the same. And by the way, why do we have 90 degrees? Well, because some of these two acute angles of the right triangle is always 90 degrees. So from 180, we subtract some of two acute angles. Either these or these doesn't really matter. And we always get 98 in the remainder. So we have a rhombus which has all the angles equal to 90 degrees, which is a square. Okay, the last problem. What condition should the quadrangle satisfy? If a new quadrangle with vertices at midpoints of each side, form parallelogram, or rhombus, or a triangle, or square. So we have some kind of a quadrangle. Now we connect midpoints. Question is, what condition should this quadrangle satisfy for this new quadrangle to be a parallelogram? The answer is no other additional condition because for any quadrangle, if you connect midpoints, you will get parallelogram. Why? Because if you consider this diagonal in a triangle, A, B, C, E, F, G, H, E, F would be a mid-segment because it connects two midpoints of sides of this triangle. And we know that the mid-segment is always parallel to the base and equal its half. But from the triangle A, C, D, G, H is also mid-segment because G is midpoint of this and H is midpoint of this, which means G, H is also parallel to A, C and equal its side, its half. So E, F and G, H are both parallel to A, C and equal its half. In our case for proving that this is parallelogram, forget about the lengths that this is one half of the diagonal. What's important is that it's parallel. Now, the same thing would be with this diagonal. For this diagonal, E, H is a mid-segment in a triangle A, B, G, which means it's parallel to B, G and equal its half, but we will use it later on. B, C, G triangle has F, G as a mid-segment, which means F, G is also parallel to B. So these are always parallel, which means that the sub-problem A, what conditions should a quadrangle satisfy to have these midpoints connected making a parallelogram, no other condition. Any quadrangle will do. Now, how about Rombus? Well, Rombus has a property that the sides are not only parallel but also equal to each other in lengths. They are combined. Now we know that two diagonals are actually twice as big as parallel to them sides. So F, G and E, H are equal to each other and equal to half of B, B. Now, G, H and E, F also are parallel and equal to each other and equal to half of A, C. So for these sides, all of them to be congruent equal in size, we have to have diagonals which are equal in size. So congruent diagonals are needed for this particular quadrangle, which is formed by midpoints, to become a Rombus. So parallel, parallelogram is always Rombus only if diagonals are congruent to each other. They are equal in lengths. How about rectangle? Okay, since again these sides are parallel to corresponding diagonals and we know that rectangle has right angle between the sides, so these angles are supposed to be right angles. For them to be right, we need this angle to be right. So if diagonals are perpendicular to each other, that makes not necessarily equal in lengths, only perpendicular to each other. That's sufficient condition for the sides of this parallelogram to be also perpendicular to each other, making all the right angles and that's why it becomes a rectangle. And finally square. Now square is Rombus and a rectangle. It has to have equal in size angles since it's a rectangle and equal in size sides because it's a Rombus. So that's what makes a square. So in this particular condition, we just have to combine whatever we have already determined for Rombuses and rectangles, which means if our diagonals are congruent equal in size to each other and perpendicular to each other, then we will have a square if we connect midpoints. Well, basically that's it. That was the last problem. Thanks for your attention and don't forget to go to Unisor.com for all additional educational material. And I would also like to encourage parents or supervising teachers to go to the site and use its ability to control the educational process of your students by enrolling them and checking their exam score. Thanks very much and good luck.