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# 数学代写非欧几何代写Non-Euclidean Geometry代考|MAT470 The Existence of Similar Figures

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## 数学代写非欧几何代写Non-Euclidean Geometry代考|The Existence of Similar Figures

The following statement is also equivalent to the Fifth Postulate and may be substituted for it, leading to the same consequences.
There exists a pair of similar triangles, i.e., triangles which are not congruent, but bave the three angles of one equal, respectively, to the three angles of the otber.

To show that this is equivalent to the Fifth Postulate, we need only show how to deduce the latter from it, since every student of Euclid knows that the use of the Postulate leads to a geometry in which similar figures exist.

Given two triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ (Fig. 7) with angles $A, B$ and $C$ equal, respectively, to angles $A^{\prime}, B^{\prime}$ and $C^{\prime}$. Let $A B$ be greater than $A^{\prime} B^{\prime}$. On $A B$ construct $A D$ equal to $A^{\prime} B^{\prime}$ and on $A C$ construct $A E$ equal to $A^{\prime} C^{\prime}$. Draw $D E$. Then triangles $A D E$ and $A^{\prime} B^{\prime} C^{\prime}$ are congruent. The reader can easily show that $A E$ is less than $A C$, for the assumption that $A E$ is greater than or equal to $A C$ leads to a contradiction. It will not be difficult now to prove that the quadrilateral $B C E D$ has the sum of its four angles equal to four right angles.

Very shortly we shall prove, ${ }^8$ without the use of the Fifth Postulate or its equivalent, that (a) the sum of the angles of a triangle can never be greater than two right angles, provided the straight line is assumed to be infinite, and (b) if one triangle has the sum of its angles equal to two right angles, then the sum of the angles of every triangle is equal to two right angles. By the use of these facts, our proof is easily completed.

By drawing $B E$, two triangles, $B D E$ and $B C E$, are formed. The angle-sum for nelther is greater than two right angles; if the anglesum for either were less than two right angles, that for the other would have to be greater. We conclude that the sum of the angles for each triangle is equal to two right angles and that the same is then true for every triangle.

## 数学代写非欧几何代写Non-Euclidean Geometry代考|Equidistant Straight Lines

Another noteworthy substitute is the following:
There exists a pair of straight lines everywbere equally distant from one anotber.

Once the Fifth Postulate is adopted, this statement follows, for then all parallels have this property of being everywhere equally distant. If the above statement is postulated, we can easily deduce the Fifth Postulate by first proving that there exists a triangle with the sum of its angles equal to two right angles.

Let $A B$ and $C D$ (Fig. 8) be the two lines everywhere equally distant. From any two points $O$ and $Q$ on $C D$ draw $O P$ and $Q R$ perpendicular to $A B$, and from any point $S$ on $A B$ draw $S T$ perpendicular to $C D$. By hypothesis $O P, Q R$ and $S T$ are equal. Since right triangles $O P S$ and $O T S$ are congruent,
Similarly
$\angle P S O=\angle T O S$.
$\angle R S Q=\angle T Q S$.
It follows that the sum of the angles of triangle $O S Q$ is equal to two right angles.

∠磷小号○=∠吨○小号.
∠R小号问=∠吨问小号.

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