Posted on Categories:Complex Geometry, 复几何, 平面几何, 数学代写

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## 数学代写|复几何代写Complex Geometry代考|Varignon’s Theorem

Varignon’s Theorem is a statement in Euclidean geometry by Pierre Varignon that was first published in 1731. It deals with the construction of a particular parallelogram (Varignon parallelogram) from an arbitrary quadrangle (another name for a quadrilateral).
The midpoints of the sides of an arbitrary quadrangle form a parallelogram. If the quadrangle is convex, then the area of the parallelogram is half as large as the area of the quadrangle.
You will prove the first part of the theorem by solving Problem 35
Problem 35. Prove that if consecutive midpoints of all sides of a quadrilateral are connected, that they form a parallelogram.

Solution. I have watched many students trying to solve this problem. Many of them draw a nice, accurate picture of the problem. From the picture, they make a conclusion: “It is always a parallelogram.” Yes, an accurate picture is $50 \%$ of a successful solution in geometry, but it is not the solution itself.

## 数学代写|复几何代写Complex Geometry代考|Trapezoids

A trapezoid is a convex quadrilateral with at least one pair of parallel sides. Any pair of angles adjacent to a parallel side sums to $180^{\circ}$ since they are supplementary. The parallel sides are bases of the trapezoid. A midline segment (midsegment) of a trapezoid connects the midpoints of nonparallel sides and is parallel to the bases.

Trapezoid Midsegment Length Theorem. The length of a midsegment is equal to $\frac{1}{2}(a+b)$, where $a$ and $b$ are the bases of a trapezoid.

So, for a trapezoid with bases 4 and 10, irrespective of the lengths of other sides, the midline segment equals $(4+10) / 2=7$.

We will give the proof of the Trapezoid Midsegment Length Theorem in Problem 41.

Problem 41. A trapezoid $A B C D$ with bases $A D=a$ and $B C=b$ is given. $M$ is the midpoint of $A B, N$ is the midpoint of $C D$. Prove that $M N=(a+b) / 2$.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Complex Geometry, 复几何, 平面几何, 数学代写

## avatest™帮您通过考试

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## 数学代写|复几何代写Complex Geometry代考|Angle Bisector and its Properties

Theorem of the Three Angle Bisectors. In a triangle all angle bisectors intersect at one point. This point is the center of the circle inscribed into the triangle.
One of the proofs of this statement can be done easily with the use of Ceva’s Theorem and knowledge of some properties of angle bisectors that are given below. However, just for demonstration, we sketch
Figure $1.45$ with the use of Geometry Sketch Pad and obtain the following:

Triangle Angle Bisector Theorem. A bisector of an angle of a triangle divides the opposite side of the triangle into segments, which are proportional to the adjacent sides of the triangle.
The following formulas are valid.
$$\frac{b}{c}=\frac{a_b}{a_c}, \quad \frac{a}{b}=\frac{c_a}{c_b}, \quad \frac{c}{a}=\frac{b_c}{b_a}$$

## 数学代写|复几何代写Complex Geometry代考|Median, Bisector and Height from a Vertex

Here are formulas for the bisector, $l_a$, median, $m_a$, and height, $h_a$ dropped from the same vertex to the opposite side, $a$, of triangle $A B C$ (Fig. 1.48):
\begin{aligned} l_a &=\frac{2 b c \cdot \cos \left(\frac{\angle A}{2}\right)}{b+c} \ l_a^2 &=b c-a_b \cdot a_c \ m_a &=\frac{1}{2} \sqrt{2 b^2+2 c^2-a^2} \ h_a &=\frac{b c}{a} \cdot \sin (\angle A) \end{aligned}
Note. By analogy, you can rewrite these with respect to other sides.

Consider further the triangle $A B C$ of Fig. $1.48$ where $A B=c, C B=a$, and $A C=b . M$ is the midpoint of $B C$ and $m_a=A M$ is a median. If $m \angle B A D=m \angle D A C$, then $l_a=A D$ is the bisector of $\angle B A C$. Let $B D=a_c, D C=a_b, A G \perp B C$ and $h_a=A G$ is the height of triangle $A B C$.

## 数学代写|复几何代写Complex Geometry代考|Angle Bisector and its Properties

$$\frac{b}{c}=\frac{a_b}{a_c}, \quad \frac{a}{b}=\frac{c_a}{c_b}, \quad \frac{c}{a}=\frac{b_c}{b_a}$$

## 数学代写|复几何代写Complex Geometry代考|Median, Bisector and Height from a Vertex

$$l_a=\frac{2 b c \cdot \cos \left(\frac{\angle A}{2}\right)}{b+c} l_a^2 \quad=b c-a_b \cdot a_c m_a=\frac{1}{2} \sqrt{2 b^2+2 c^2-a^2} h_a \quad=\frac{b c}{a} \cdot \sin (\angle A)$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Complex Geometry, 复几何, 平面几何, 数学代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|复几何代写Complex Geometry代考|Law of Cosines and Law of Sines

The Law of Cosines is the relationship between the sides and the angles of a triangle: $a^2=b^2+c^2-2 b c \cdot \cos (\angle A)$, where $\angle A$ is opposite of side $a$. Students usually learn this formula at the end of a geometry course but don’t understand how to apply it until taking trigonometry. Nevertheless, this formula is very powerful. In addition, the Pythagorean Theorem is a particular case of the Law of Cosines. Let us now solve Problem 8 :

Problem 8. A triangle with sides of length 5, 12, and 13 is given. Find the angle that is opposite the biggest side.

Solution. First, we will draw an accurate picture and put all known information on it so it will help us to calculate the measure of angle $C$ (Fig. 1.15).

Let us apply the Law of Cosines to the triangle:
\begin{aligned} &13^2=5^2+12^2-2 \cdot 5 \cdot 12 \cdot \cos (\angle C) \ &0=\cos (\angle C), \quad \text { then } \quad m \angle C=90^{\circ} \end{aligned}
We found that triangle $A B C$ is a right triangle.
Let us prove that the Pythagorean Theorem is a particular case of the Law of Cosines. For this purpose we will write the Law of Cosines for the cosine of angle $C$ using the same picture:
\begin{aligned} &c^2=a^2+b^2-2 a b \cos (\angle C) \ &\cos (\angle C)=\frac{a^2+b^2-c^2}{2 a b} \end{aligned}

## 数学代写|复几何代写Complex Geometry代考|Similar Triangles

Two triangles are called similar if their corresponding angles are equal and the ratio of corresponding sides is the same. Similar triangles or, in general, any similar figures have a similar shape. You probably remember that two triangles are similar to each other by two angles $(A A)$, by two sides and the included angle $(S A S)$, and by three sides (SSS). Especially important is the fact that in similar triangles the ratio of corresponding sides, medians, heights, and bisectors equals $k$, the coefficient of similitude. The ratio of the areas of similar triangles equals $k^2$, the square of the coefficient of similitude.

The following picture will illustrate one way of constructing similar triangles by hand, with $k>1$ (magnification) and $k<1$ (making the image smaller). This is based on Homothetic Transformation or such transformation at which any point of the original (pre-image), its image and the center of homothety lie on the same line. Using homothety, different similar figures can be constructed, so all corresponding angles are equal.

## 数学代写|复几何代写Complex Geometry代考|Law of Cosines and Law of Sines

$$13^2=5^2+12^2-2 \cdot 5 \cdot 12 \cdot \cos (\angle C) \quad 0=\cos (\angle C) \text {, then } m \angle C=90^{\circ}$$

$$c^2=a^2+b^2-2 a b \cos (\angle C) \quad \cos (\angle C)=\frac{a^2+b^2-c^2}{2 a b}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。