$ABCD$ is a parallelogram and $O$ is any point. The parallelogram $OAEB,OBFC,OCGD,ODHA$ are completed. show that $EFGH$ is a parallelogram.
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In the following figure, $ABCD$ is a parallelogram, and $O$ is any point. Parallelograms $OAEB, OBFC, OCGD, ODHA$ are completed. Prove that $EFGH$ is a parallelogram.   We can obtain a fairly trivial proof using affine geometry. As $OAEB, OBFC, OCGD, ODHA$ are parallelograms,   $$\vec{A} + \vec{B} = \vec{O} + \vec{E} \implies \vec{A}  + \vec{B} - \vec{O} = \vec{E} \tag1$$  $$\vec{C} + \vec{B} - \vec{O} = \vec{F}\tag2$$  $$\vec{D} + \vec{C} - \vec{O} = \vec{G} \tag3$$  $$\vec{D} + \vec{A} - \vec{O} = \vec{H}\tag4$$  Adding $(1)$ to $(3)$ and $(2)$ to $(4)$, we get,   $$\vec{E} + \vec{G} = \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag5$$  $$\vec{F} + \vec{H} =  \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag6$$  Clearly, $(5)$ and $(6)$ are equal, therefore,   $$\vec{E} + \vec{G} = \vec{F} + \vec{H}$$  Therefore, $EFGH$ is a parallelogram. Can somebody give an elementary proof using Euclidean geometry? Also, I noticed that in my proof, nowhere did I use the fact that $ABCD$ is a parallelogram, but constructing an example, it was quickly clear that the result stated does not generalize to all quadrilaterals. How come? Is my proof incorrect?
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I've noticed something while trying to prove the properties of parallel lines, and the properties that a triangle has 180 degrees. To prove the properties of parallel lines, such as alternate angles, you need to use the property that a triangle has 180 degrees. To prove a triangle has 180 degrees however, you need to use the properties of parallel lines. This really bothers me because of how circular it is. They are both reliant on each other to be true, and do not logically show, without being reliant on each other, why triangles have 180 degrees, and why parallel line properties are true.   So what I hoping for here is a way to prove parallel line properties without using the fact that a triangle has 180 degrees, or a way to prove triangles have 180 degrees without using parallel line properties. This way, things will be logical to me, and make sense.   Thanks in advance.
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We consider in $\mathbb{R}^2$ the set of points $$\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}$$  Let $\Omega$ a rectangular structure consisting of the heads $\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}$  Let the two polygons $\sum_1 = \{M_4,M_5,M_2,M_1\}$ and $\sum_2 = \{M_5,M_6,M_3,M_2\}$   let the mesh $\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}$ such $$Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}$$  1- Prove that $(e_1,\sum_1,Q_1)$ and $(e_2,\sum_2,Q_1)$ are two finite elements of Lagrange.  2- Let $V$ the finite-dimensional space that corresponds to the mesh. Prove that $V \nsubseteqq H^1(\Omega)$  My problem is to prove 2.
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This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
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Let $H&lt;S_{n}$ and suppose $H$ is not contained in $A_{n}$. Write $H$ as $$H=E\cup O$$ where $E$ and $O$ represent the sets of even and odd elements, respectively. Let $E=\{{\alpha_1,...,\alpha_n}\}$ and $O=\{{\beta_1,...,\beta_m}\}$. Since $O$ is non-empty, it contains at least one element, say $\beta_1$. The same goes for $E$ because $e\in E$. Our goal is to prove $m=n$. The products $\beta_1\alpha_i$, $i=1,...n$, are all distinct elements in $O$, and it follows that $O$ has at least $n$ elements, i.e. $m\geq n$. In the case of strict inequality, $m&gt;n$, add at least one extra odd element ($\beta_{n+1}$) to the list $$\beta_1\alpha_1,...,\beta_1\alpha_n,\beta_{n+1}$$  This is where I find myself stuck. Any help appreciated. Can this route work?
86572
How should I approach obvious proofs like these    I have been trying but couldn't find an elegant way to work these. Any help is highly appreciated ! Especially looking for a nice proof/hint for first problem
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Can someone of you give me a solution for this?     Let $N\in \mathbb N$.   a) We define the map $\mathfrak F:(\mathbb C^N, ||\cdot||_2)\to(\mathbb C^N, ||\cdot||_2)$ by $$(\mathfrak F(x))_k := \frac{1}{\sqrt N}\sum_{j=1}^Nx_je^{2\pi i\frac{(j-1)(k-1)}{N}} \quad \forall k\in \{1,...,N\}$$    Show that $\mathfrak F$ is norm-preserving, i.e. $$||(\mathfrak F(x))_k||_2 = ||x||_2 \quad \forall k\in \{1,...,N\}$$    b) For $n,m \in \{1,2,...,N\}$ we define the entries of $M\in \mathbb C^{N\times N}$ by $$M_{nm} :=  \begin{cases} \frac{N+1}{2}, \quad \quad n=m \\ \frac{1}{e^{2\pi i \frac{m-n}{N}}-1}, \,\,n \neq m \end{cases}$$   Show that $1, 2, 3, ..., N$ are the Eigenvalues of M.
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Given four points $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$,$(x_4,y_4)$. How can we efficiently test them to make sure whether they are vertices of a parallelogram?  I think evaluating and comparing the distance between two points seems to be inefficient. What do you think? Do you have any better method?
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I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.
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Below the parallelogram is obtained from square by stretching the top side while fixing the bottom. Since area of parallelogram is base times height, both square and parallelogram have the same area.    This is true no matter how far I stretch the top side.    In below figure it is easy to see why both areas are same.   But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?
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I am hoping that someone can help me see understand a proof that the sum of angles of a hyperbolic triangle $\triangle ABC$ is strictly less than $\pi$.  I want to stick to the upper-half-plane model $\mathbb H:=\{z\in \mathbb C : Im (z)&gt;0\}$ of hyperbolic geometry where lines are either vertical lines or semicircles centered on (but excluding) the $x$-axis.  To simply things, by using an LFT we can take one vertex $A$ to be the point $i$ and another $B$ to be $(0,k)\in \mathbb H$ for some $k&gt;0$.    Now, call the third vertex $C$. We have the other two sides $AC$ and $BC$ as arcs of the circles $C_1 =(x-a)^2+y^2=r_1^2$ and $C_2=(x-b)^2+y^2=r_2^2$, respectively, so that $x^2+1=r_1^2$ and $x^2+k^2=r_2^2$.  In my class notes, there is a brief mention of how this simplification can help us with the proof. However, I do not see at all how this helps us get the result that $\triangle ABC$ sums to less than $\pi$. Therefore, any help is very appreciated.  P.S., My knowledge is elementary, so I am hoping for a proof using basic tools.
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Can someone please check my proof and my definitions.  Let $A \in \mathbb{R}^{n \times m}$ be my matrix.  The left null space of $A$ is written as,  $\mathcal{N}(A^\top) = \{x \in \mathbb{R}^n| A^\top x = 0\}$  The orthogonal complement of the column space $\mathcal{C}(A)$ is written as,  $\mathcal{C}(A)^\perp = \{x \in \mathbb{R}^n | x^\top y = 0, \forall y \in \mathcal{C}(A)\}$     We want to show that $\mathcal{N}(A^\top)  = \mathcal{C}(A)^\perp $   First, we show, $\mathcal{N}(A^\top) \subseteq \mathcal{C}(A)^\perp$  Let $x \in \mathcal{N}(A^\top)$, then $A^\top x = 0 \implies x^\top A = 0^\top \implies x^\top Av= 0^\top v, \forall v \in \mathcal{C}(A) \implies x^\top y = 0 , y = Av$, $\implies x \in C(A)^\perp$.  Next, we show, $\mathcal{N}(A^\top) \supseteq \mathcal{C}(A)^\perp$  Let $x \in C(A)^\perp$, then $x^\top y = 0$, forall $y \in C(A)$. But $y = Av, \forall v \in \mathbb{R}^n$. Hence, $x^\top y = x^\top Av = v^\top A^\top x.$ For all $v \neq 0, A^\top x = 0$, hence $x \in \mathcal{N}(A^\top)$.    I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
121292
Points $P$, $Q$, $R$ and $S$ are chosen on the sides of parallelogram $ABCD$, so that $P$ is on line $AB$, $Q$ is on line $BC$, $R$ is on line $CD$, $S$ is on line $DA$, and $AP=BQ=CR=DS=\frac{1}{3}(AB)$. Compute the ratio of the area of $PQRS$ to the area of $ABCD$.
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I've seen how to visualize the formula for getting the area of a parallelogram. The first picture shows 2 ways which give the same result of Area = base * height.   (link to first pic)  However, the first time I thought about it before being show the actual formula, I thought it was Area = side * side.   (link to second pic)  The 2 sides of the parallelogram are 2 vectors, x and y.  If you place copies of vector y length of x times along the vector x, and then you add all lengths of vector y's (which is the same as length of vector y * length of vector x), it seems like you also get the area of the parallelogram.   I know that you'll end up with a different result since the height is always less than the length of the vector y (unless they were parallel) but the method in the second picture seems correct.   I also haven't seen the second picture's method being taught online so I'm guessing it's actually incorrect. What I want to know is what makes it incorrect because it really seems like it's correct despite giving a different answer from the usual base * height formula.  Thanks in advance for any help.
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When using the definition and properties of the inner product, we get the parallelogram law: $$\:\,\|x+y\|^2 =\: \langle x+y, x+y\rangle= \langle x, x\rangle + \langle x, y\rangle +\langle y, x\rangle +\langle y, y\rangle\,,\\ \|x-y\|^2 =\:\langle x-y, x-y\rangle= \langle x, x\rangle - \langle x, y\rangle -\langle y, x\rangle +\langle y, y\rangle $$ Adding these two expressions: $$\|x+y\|^2+\|x-y\|^2 \:=\: 2\langle x, x\rangle + 2\langle y, y\rangle \:=\: 2\|x\|^2+2\|y\|^2$$ as required. But the above does not simplify to Pythagoras' theorem $\|x+y\|^2=\|x\|^2+\|y\|^2$ if $x$ and $y$ are orthogonal. How can we get Pythagoras from the parallelogram law?
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Can anyone give me a vectorial solution to the following problem:     Prove that if each pair of opposite edges of the tetrahedron $ABCD$ is perpendicular (that is, $AB \perp CD$ and $AC \perp BD$ and $AD \perp BC$), then the heights of the tetrahedron are concurrent.   Here, the heights (also known as the altitudes) of tetrahedron $ABCD$ are the perpendicular from $A$ to the plane $BCD$, and three other similarly defined perpendiculars.  Tetrahedra satisfying the condition of this problem are called , and this appears to be a known result.
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Let $\alpha:I\to\Bbb R^3$ be a parameterized curve and let $v\in \Bbb R^3$ be a fixed vector. Assume that $\alpha'(t)$ is orthogonal to $v$ for all $t\in I$, and that $\alpha(0)$ is also orthogonal to $v$. Prove that $\alpha(t)$ is orthogonal to $v,\forall t\in I$.   I have been having trouble solving this problem, and I have written it in math as follows:  $$v_1x'(t)+v_2y'(t)+v_3z'(t)=0=v_1x(0)+v_2y(0)+v_3z(0)$$  and using this I want:  $$v_1x(t)+v_2y(t)+v_3z(t)=0$$  Can I have a hint please?