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Transcriber's Note:The Index has been updated to correspond with the page numbers of this edition. Even though the The author hoped that “few typos have gone unnoticed," but several were found and have been Please correct and list it at the end of the text.

This edition of the Elements of Euclid, created at the request of leaders from some of the top colleges and schools in Ireland, aims to address a significant need felt by teachers today— the creation of a work that not only presents the unmatched original in its entirety but also incorporates modern concepts and developments in the geometry covered by the Elements. A brief look at the work will reveal that the Editor has gone much further in this area than any of his predecessors; it contains not only more material than any of theirs that he knows of, but also much that is unique, which he believes hasn’t appeared in any previous works on the topic. The significant expansion of geometric methods in recent times has made this work essential for students, helping them not only to read effectively but also to grasp those contemporary mathematical texts that demand a solid understanding of Elementary Geometry, making this book the best introduction to it.

In putting this work together, the Editor received invaluable help from the late Rev. Professor Townsend, s.f.t.c.d. The book was rewritten and significantly changed based on his suggestions, and it owes much of its merit to that distinguished Geometer.

The Questions for Examination in the early part of the First Book are meant to serve as examples that the teacher should use throughout the whole work. Anyone who has experience in teaching understands the significance of these exams in teaching Basic Geometry.

The Exercises, which number over eight hundred, have all been carefully chosen. The ones included in each Book are meant to apply Euclid’s Propositions. Most of them are quite basic and can be considered common knowledge, as nearly all have appeared in earlier collections. The Exercises at the end of each Book are more advanced; several come from the late Professor Townsend, some are original, and many have been sourced from two significant French works—Catan's Théorèmes et Problèmes de Géométrie Elélémentaire, and the Characteristicé de Géomunchanged, by Rouché and De Comberouse.

The second edition has been completely updated and significantly expanded. The new content includes several alternative proofs, important exam questions for each of the books, an explanation of the ratio of incommensurable quantities, the first twenty-one propositions of Book XI., and an Appendix on the properties of the Prism, Pyramids, Cylinder, Sphere, and Cone.

The current edition has been thoroughly read, and it is hoped that only a few typos have slipped through unnoticed.

The Editor is pleased to see from the quick sale of previous editions (each 3000 copies) of his Book, and its widespread use in schools, that it is likely to achieve the two goals it was written for: to provide students with a Manual that gives them a solid understanding of the timeless work of the great Greek Geometer, and to introduce them to some of the key concepts and advancements in today's Geometry.

Geometry is the science of shaped space. Shaped space can have one, two, or three dimensions, depending on whether it consists of lines, surfaces, or solids. The edges of solids are surfaces; the edges of surfaces are lines; and the edges of lines are points. Therefore, Geometry focuses on studying the properties of solids, surfaces, and the shapes created on surfaces. The simplest surface is the plane, and the branch of Geometry that deals with the lines and curves on a plane is called Plane Geometry; the branch that explores the properties of solids, curved surfaces, and shapes drawn on curved surfaces is known as 3D Geometry. The simplest lines that can be drawn on a plane are the straight line and circle, and examining the properties of the point, straight line, and circle serves as an introduction to Geometry, which includes a vast and significant section. This is the area of Geometry discussed in the oldest mathematical book known, Euclid’s Components, which is the focus of this volume. The conic sections and other curves that can be drawn on a plane represent special branches and complete the divisions of this, the most extensive of all sciences. The student will find in Chasles’ Aperçu History a valuable history of the origins and development of Geometry methods.

In this work, when figures aren't provided, students should create them based on the given instructions. The Propositions of Euclid will be printed in a larger font and will be referenced by Roman numerals in brackets. For example, [III. xxxii.] will indicate the 32nd Proposition of the 3rd Book. The Book number will only be mentioned if it's different from the one being referenced. Each Proposition will present both a general and a specific statement. By skipping the letters in parentheses, you get the general statement, and by reading them, you get the specific one. The annotations will be in smaller font. The following symbols will be used in them:—

In addition to these, we will use the standard symbols +, −, &c. from Algebra, as well as the congruence sign, which is ≡. This symbol was introduced by the renowned Gauss.

iA point is something that has a position but no size.

A geometric shape with three dimensions—length, width, and thickness—is a solid, while something with two dimensions, like length and width, is a surface. Something with only one dimension is a line. However, a point is neither a solid, surface, nor line; thus, it has no dimensions, meaning it has no length, width, or thickness.

ii. A line is length with no width.

A line is a space with one dimension. If it had any width, no matter how small, it would be a space with two dimensions; and if it also had any depth, it would be a space with three dimensions; A line has no width or depth.

iii.The points where lines meet and their ends are called points.

i v.A line that is equally distant between its endpoints is called a straight or right line, like AB.

If a point moves without changing its direction, it will create a straight line. The direction in which __A_TAG_PLACEHOLDER_0__ A point that moves is called its “senseIf the moving point keeps changing its direction, it will form a curve; therefore, it follows that only one straight line can be drawn between two points. The The following illustration is by Professor Henrici."If we attach a weight to a string, the string gets stretched, and we say it is straight, meaning it has taken on a specific shape. If we imagine removing all thickness from this string, we understand the concept of the most basic line, which we refer to as straight line.

vA surface is something that has length and width.

A surface is a two-dimensional area. It has no thickness, because if it had even the smallest amount, it would become a 3D space.

v.i.When a surface is shaped so that the straight line connecting any two points on it lies completely within the surface, it is called a airplane.

A plane is entirely flat and level, similar to the surface of still water or a smooth floor.Newcomb.

vi.Any combination of points, lines, or both in a plane is called a airplanefigure. If a figure consists only of points, it's called a stigmatizedfigure; if it consists only of straight lines, it's called a straight linefigure.

viii.Points that are on the same straight line are called in a straight line points. A shape made up of collinear points is called a row of points.

ix.The angle formed by two straight lines that extend from a single point in different directions is called a straightangle.

x.The two lines are called the legs, and the point where they meet is called the vertex of the angle.

A straight line drawn from the vertex and rotating around it in the plane of the angle, transitioning from being aligned with one side to being aligned with the other, is said to rotate through the __A_TAG_PLACEHOLDER_0__. An angle increases as the amount of rotation increases. Additionally, since the line can rotate between positions in two different ways, two angles are formed by drawing two lines. from a location.

So ifAB,ACThe legs allow a line to move from its position.ABto the roleACin the two ways indicated by the arrows. The smaller angle created is understood to be the angle formed by the lines. The bigger angle, referred to as are-entrantangle, rarely appears in the "Elements."

xi.Angle Designation.—A specific angle in a figure is represented by three letters, such as BAC, where the middle letter, A, is at the vertex, and the other two are along the sides. The angle is then read as BAC.

xii.The angle created by combining two or more angles is called their total. So, the sum of the two angles ABC and PQR is the angle ABBelow is a short piece of text (5 words or fewer). Modernize it into contemporary English if there's enough context, but do not add or omit any information. If context is insufficient, return it unchanged. Do not add commentary, and do not modify any placeholders. If you see placeholders of the form __A_TAG_PLACEHOLDER_x__, you must keep them exactly as-is so they can be replaced with links.R,

formed by applying the side QP to the side BC, so that the vertex Q lands on the vertex B, and the side QR code is positioned on the opposite side of BC from BA.

xiii.When the sum of two angles BAC and CAD (Computer-Aided Design) is such that the sides BA and AD form a straight line, they are called supplements of each other.

When one line rests on another, the two angles it forms on the same side of the line it rests on are supplements of each other.

xiv. When one line stands on another and creates equal angles on both sides, each of those angles is called a right angle, and the line that stands on the other is referred to as a at right angles to it.

A right angle is equal to its supplement.

xv. An sharp angle is one that is less than a right angle, like A.

xvi. An dull angle is one that is larger than a right angle, like BAC.

The supplement the complement of an acute angle is obtuse, and conversely, the supplement of an obtuse angle is sharp.

x v i i. When the total of two angles is a right angle, each angle is referred to as the complementary of the other. So, if angle BAC is a right angle, the angles BAD and DAC are complements of each other.

xviii. Three or more straight lines that intersect at the same point are referred to as simultaneous lines.

xix. A system of more than three lines that intersect at a point is called a pencil of lines. Each line in this pencil is referred to as a ray, and the shared point through which all the rays pass is known as the vertex.

xx. A triangle is a shape made up of three straight lines connected at their ends. The three lines are known as its sides.

xxi. A triangle with all three sides different lengths is called a scalene, like A; a triangle with two sides the same length is called isosceles, like B; and a triangle with all sides the same length is called equilateral triangle, like C.

xxii. A right angletriangle is one that has a right angle, like D. The side opposite the right angle is called the hypotenuse.

xxiii. An obtusetriangle is one that has one of its angles greater than 90 degrees, as E.

xxiv. An acute-angledtriangle is one that has all three angles less than 90 degrees, as F.

xxv. An outsideangle of a triangle is created by one side and the extension of another side.

A triangle has six exterior angles, and each exterior angle is the supplement of the adjacent interior angle.

xxvi. A straight figure enclosed by more than three straight lines is generally called a polygon.

xxvii. A polygon is considered curved outwards when it doesn't have any inward angles.

xxviii. A four-sided shape is called a quad.

xxix. A quadrilateral with four equal sides is called a cough drop.

xxx. A shape with a right angle is called a square.

xxxi. A shape with five sides is called a pentagon; a shape with six sides is a hexagon, and so on.

xxxii. A circle is a flat shape created by a curved line known as the circumference, where all straight lines drawn from a specific point inside the shape to the circumference are equal in length. This point is called the center.

xxxiii. A radius of a circle is any line drawn straight from the center to the edge, like CD.

xxxiv. A diameter of a circle is a straight line drawn through the center and extending in both directions to the edge, like AB.

According to the definition of a circle, it can be inferred that the trajectory of a moving point in a plane A circle is defined as the set of points that remain at a constant distance from a fixed point; additionally, any pointPin the A plane is either inside, outside, or on the boundary of a circle based on its distance from the center. isfewerthan,greaterthan, orequalsto the radius.

Let it be known that—

i. You can draw a straight line from any point to any other point.

When we consider a straight line that is defined by two fixed points at either end, this segment is called a __A_TAG_PLACEHOLDER_0__.line segment.

ii. A straight line can be extended indefinitely in a straight path.

Any straight line can extend infinitely in either direction or both. In these cases, it's called anindeterminateline. Based on this principle, a finite straight line can be thought of as being extended at any time into an endless straight line.

iii. You can draw a circle from any center and set its radius to any length from that center.

If there are 2 pointsAandB, and if we use tools like a ruler and a pen to draw a line fromAtoBThis line will definitely have some inconsistencies, along with some width and thickness. So, it won't be a true geometric line, no matter how similar it may look. That's why Euclid says we should draw a straight line from __A_TAG_PLACEHOLDER_0__. one point to another. If it could be done perfectly, he wouldn't need to ask us to accept it as a given. The same logic applies to the other postulates. It's also It's important to note that Euclid never assumes that anything can be done that can't be constructed geometrically using other problems or earlier postulates.

i. If two things are equal to the same thing, or to things that are equal to each other, then they are equal to each other.

So, if we have three things, and the first and second are each equal to the third, we We can conclude from this principle that the first is equal to the second. This principle applies to all kinds of measurements. The same applies to Axioms.ii.,iii.iv.v.vi.vii.,ix.; butviii.x.xi.,xii., are strictly geometric design.

ii. If you add equal amounts to equal amounts, the totals will be equal.

iii. If you take the same amount from both sides, the leftovers will be the same.

iv. If you add equal amounts to unequal ones, the totals will be unequal.

v. If you take away equal amounts from unequal amounts, the leftovers will also be unequal.

vi. Doubles of the same magnitude are equal.

vi. The halves of equal sizes are equal.

viii. Magnitudes that can be aligned are equal.

Putting one geometric shape over another, like a line on a line, a triangle on a __A_TAG_PLACEHOLDER_0__, a triangle, or a circle on a circle, etc., is calledsuperpositionThe concept of superposition in geometry is justmindset, meaning we envision one shape positioned over the other; and then, if we can demonstrate that they align perfectly, we conclude, based on the current principle, that they are equal. Superposition includes the following principle that Euclid often uses without directly mentioning it:—“Any figure can be moved from one position to another without altering its shape or size.

ix. The whole is greater than the sum of its parts.

This principle is included in the following, which is a more detailed statement:—

ix′. The whole is equal to the total of all its parts.

x. Two straight lines cannot form a closed area.

This means, "If two straight lines share two points, they __A_TAG_PLACEHOLDER_0__. “point in the same direction,” which means they are essentially the same line, and this holds true even when one of the points are at infinity.

xi. All right angles are the same as each other.

This can be shown like this: Let there be two straight lines__A_TAG_PLACEHOLDER_0__ABandCD, along with two intersecting lines to them, specifically,EFandGHIf we alignABandCDby overlapping them, so that pointE matches withG, then since a right angle is equal to its supplement, the lineEFmust align withGH. So, the angleAEFis equal toCGH.

xii. If two straight lines (AB,CD) intersect a third line (AC) in such a way that the sum of the two interior angles (BACACD) on the same side is less than two right angles, then these lines, if extended, will meet at some limited distance.

This principle is the opposite of Prop. xvii., Book I.

Principles.—"Elements of human reasoning," according to Dugald Stewart, are certain general statements that are obviously true and so fundamental that they can't be derived from any more basic statements; in other words, they can't be proven. "The two sides of a triangle are greater than the third" might seem obvious, but it's not an axiom since it can be proven from other statements; however, we can't prove the statement that "things equal to the same thing are equal to each other," and because it is self-evident, it is an axiom.

Suggestions that aren't axioms are characteristics of figures derived from reasoning processes. They are categorized into theorems and problems.

A Theorem is a formal statement about a property that can be proven based on established propositions. These propositions can be either theorems or axioms. A theorem has two parts: the theory, which is what is assumed, and the conclusion, which is what is claimed to follow from it. Therefore, in a typical theorem,

the hypothesis is that Xis Y, and the conclusion is that Zis W.

Reverse Theorems.—Two theorems are considered converse when the premise of one is the conclusion of the other. Therefore, the converse of the theorem (i.) is—

From the two theorems (i) and (i i.) we can derive two others, known as their contrapositives. So, the contrapositive

The theorem (iv.) is known as the reverse of (i.) and (iii.) is the obverse of (ii.).

A Issue is a statement suggesting that something should be accomplished, like drawing a line or creating a shape, under specific conditions.

The Solution of a problem is the way of building that achieves the desired outcome.

The Demo is the evidence, in the case of a theorem, that the conclusion comes from the hypothesis; and in the case of a problem, that the construction achieves the intended goal.

The Pronunciation of a problem has two parts: the data, which are the things that are assumed to be provided, and the requested, which are the things that need to be accomplished.

Assumptionsare the basic components of geometric construction and have the same role in relation to problems as axioms do to theorems.

A Additional point is a conclusion or reasoning drawn from a statement.

A Lemma is a supporting statement needed to prove a main statement.

A Secant line or Cross-cutting is a line that intersects a set of lines, a circle, or any other geometric shape.

Aligned figures are those that can be made to overlap perfectly. They have the same shape and size but are positioned differently. Therefore, according to Axiom viii., the corresponding parts of congruent figures are congruent, and these figures are equal in all aspects.

Identity Rule.—This principle is sometimes stated as follows:—“If there is only one Xand one Y, then if X is Y, it logically follows that Y is X.”—Syllabus.

Sol.—Using Aas the center and ABas the radius, draw the circle BCD(Post. iii.). With B as the center and BA as the radius, draw the circle ACE, which intersects the first circle at C. Connect CA and CB (Post. i). Then, ABCis the equilateral triangle you need.

Dem.—Since Ais the center of the circle BCD, ACis equal to AB (Def. xxxii.). Also, since Bis the center of the circle ACE, BCis equal to BA. Therefore, we have proven this.

But things that are equal to the same thing are equal to each other (Axiom i.); therefore, AC is equal to BC; so the three lines AB, Before Christ, and CA are equal to each other. Thus, the triangle ABC is equilateral (Def. xxi.); and it is constructed on the given line AB, which was what we needed to accomplish.

What is the data in this proposition?

What is the question?

3.What is a straight line with a specific length?

4.What is the opposite of finite?

5.In which part of the construction is the third postulate referenced, and what is its purpose? Where Is this the first postulate referenced?

6.Where is the first axiom referenced?

7.How is the definition of a circle applied? What is a circle, exactly?

What’s an equilateral triangle?

The following exercises should be done after the student has mastered the First Book:

1.If the linesAFandBFare connected, the formACBFis a diamond.

2.IfABis extended toDandE, the trianglesCDFandCEFare equilateral.

3.IfCAandCBare created to connect with the groups again atGandH, the topicsG,F,Harecollinear, and the triangleGCHis equilateral.

4. If CF is joined, CF² = 3AB².

5.Describe a circle in the area.ACB, restricted by the lineABand the two circles.

Sol.—Connect AB (Post. i.); on AB draw the equilateral triangle ABD[i.]. With B as the center, and BC as the radius, draw the circle ECH (Post iii.). Extend DB to intersect the circle ECH at E (Post. i i.). With D as the center, and DE as the radius, draw the circle EFG (Post. iii.). Extend DA to intersect this circle at F. AFis equivalent to BC.

Dem.—Since D is the center of the circle EFG, DF is equal to DE (Def. xxxii.). And since Dab is an equilateral triangle, DA is equal to DB (Def. xxi.). So, we have

and taking the latter from the former, the remainder AF is equal to the remainder BE (Axiom iii.). Again, since B is the center of the circle ECH, BC is equal to Be; and we have proven that AF is equal to Be; and things that are equal to the same thing are equal to each other (Axiom i.). Hence, AF is equal to BCE. So, from the given pointAthe lineAFhas been made equal toBefore Christ.

Commentators on Euclid often claim that he allows the use of the __A_TAG_PLACEHOLDER_0__.rulerandcompass. If that were the case, this Proposition wouldn’t be needed. The truth is, Euclid intended to Teach theoretical geometry, not practical geometry, and the only things he assumes are the ability to draw straight lines and create circles. If he had permitted the mechanical use of the With a ruler and compass, he could offer techniques for solving numerous problems that go beyond the scope of the "geometry of the point, line, and circle."See you later.Notes D, F at the end of this. job.

1.Solve the problem when the point __A_TAG_PLACEHOLDER_0__Ais at stakeBefore Christitself.

2.Draw a line from pointAto the lineBCthat has the same length as a specified line. Indicate how many solutions exist.

Sun.—From A, one end of AB, draw a straight line AD that is equal to C [ii.]; and using A as the center and AD as the radius, draw the circle EDF (Post. iii.) that intersects AB at E. AE will be equal to C.

Dem.—Since Ais the center of the circle EDF, AEis equal to AD (Def. xxxii.), and Cis equal to AD(const.); and things that are equal to the same thing are equal to each other (Axiom i); therefore, AEis equal to C. Thus, from ABthe longer of the two provided lines, a segment,AE, has been cut off to matchC, the shorter.

Your text appears to be incomplete. Please provide a phrase for me to modernize.What previous problem is being used to solve this?

What assumption?

3. What principle in the demonstration?

4.Show how to draw the shorter of two given lines until the entire drawn line is equal in length. to the longer version.

If two triangles(BAC,EDF) have two sides to it(BA,AC) that are equal to two sides(ED,DF) of the other, and also have the angles(A,D) If those sides are equal, then the triangles will be equal in every way—that is, their bases or third sides(BCEF) will be equal, and the angles(B,C) At the base of one will be equal to the angles.(E,F) at the the other; specifically, those angles will be the same as the angles opposite the equal sides.

Dem.—Let’s imagine triangle BAC is positioned on EDF, so that point A aligns with D, and line AB aligns with DE, while point C is on the same side of DE as F; therefore, since AB equals DE, point B must align with E. Moreover, since angle BAC is equal to angle EDF, line AC must line up with DF; and because AC equals DF (hyp.), point C must align with F; thus, we have shown that point B aligns with E. Therefore, two points on line BC coincide with two points on line EF; since two straight lines cannot enclose a space, BC must coincide with EF. Thus, the triangles are congruent in every way; thusBCequalsEF, angleBequalsE, perspectiveCequalsF, and triangleBACis the same as triangleEDF.

1.How many parts are there in the hypothesis of this Proposition?Ans.Three. Write them down.

2.How many are in the conclusion? List them.

3.What technical term refers to figures that are the same in every way except for their position?Ans.They are called congruent.

What does superposition mean?

5.What principle is used in superposition?

6.How many sides does a triangle have?Ans.Six; more specifically, three sides and three angles.

7.When you need to prove that two triangles are congruent, how many parts of one must you compare to the other? Is it shown to be equal to the corresponding parts of the other?Ans.Generally, any three parts except the three angles. This will be demonstrated in Props. viii.andxxvi., together with i.v.

8.What characteristic do two lines that share two points have according to this Proposition? They __A_TAG_PLACEHOLDER_0__ must be in the same line.

1.The line that splits the vertical angle of an isosceles triangle also divides the base. perpendicularly.

2.If two adjacent sides of a quadrilateral are equal and the diagonal divides the angle between them, If you cut them in half, their other sides will also be equal.

3.If two lines intersect at right angles and each one bisects the other, then any point on either line is equally far from the ends of the other line.

4.If you draw equilateral triangles on the sides of any triangle, the distances between the __A_TAG_PLACEHOLDER_0__ The vertices of the original triangle and the opposite vertices of the equilateral triangles will be equal. (This The proposition should be proven after the student has read Prop. xxxii.Sure, please provide the text you would like me to modernize.

The angles(ABC,ACB) at the bottom(BC) The sides of an isosceles triangle are equal to each other, and if the equal sides(AB,AC) are extended, the outer angles (DEC,ECB) below the base will be the same.

Dem.—In BDpick any point F, and from AE, which is larger, cut off AG equal to AF[iii]. Connect BG and CF (Post. i.). Since AF is equal to AG (constant), and AC is equal to AB (hypothesis), the two triangles FAC and GAB have sides FA and AC that are equal to sides GA and AB, respectively; and the angle A is shared by both triangles. Thus [i.v.] the base FC is equal to GB, the angle AFC is equal to Terms and Conditions, and the angle ACF is equal to the angle ABG.

Again, since AFis equal to AG (constant), and AB is equal to AC (hypothesis), the remaining segment, BF, is equal to CG (Axiom iii); and we have shown that FC is equal to GB, and the angle BFC is equal to the angle CGB. Therefore, the two triangles BFC and CGB have the two sides BF and FC in one equal to the two sides CG and GB in the other; and the angle BFC formed by the two sides of one is equal to the angle CGB formed by the two sides of the other. Hence [i.v.] these triangles have angle FBC equal to angle GCB, and these are the angles below the base. Also, angle FC Barcelona is equal to GBC; but the entire angle FCA has been shown to be equal to the entire angle GBA. Thus, the remaining angle ACB is equal to the remaining angle ABC, and these are the angles at the base.

Observation.—The main challenge beginners encounter with this proposition is that the two trianglesACFandABGMake sure they don't overlap. The instructor should separate these triangles, as shown in the attached diagram, and point out the corresponding parts like this:—

The student should also learn how to align one triangle with the other, so that they match perfectly. Similar examples can be demonstrated using the triangles.BFC, CGB.

Here’s a straightforward proof of this proposition. Picture the __A_TAG_PLACEHOLDER_0__△ACBspinning, without changing its shape, around the lineAC, until it reaches the other side. Let's refer to the new position as __A_TAG_PLACEHOLDER_0__.ACD; then the angleADCthe angle of the moved triangle is clearly equal to the angleABC, with which it originally matched. Additionally, the two△sBACandCAD (Computer-Aided Design)have opinionsBA, ACof one that is equal to the sidesAC,Adof the other, and the angles between them are the same; thereforei.v.the angleACBacross from the sideABis equal to the angleADC across from the sideAC; but the perspectiveADCequalsABC; soACBequals ABC.

Cor.—Every equilateral triangle is equiangular.

Def.—A line in any diagram, like__A_TAG_PLACEHOLDER_0__,ACIn the previous diagram, there is a line that, when you fold the plane of the figure along it, will have one side of the diagram align perfectly with the other. This is called an __A_TAG_PLACEHOLDER_0__.axis of symmetry of the figure.

1.Prove that the angles at the base are equal without extending the sides. Additionally, by extending __A_TAG_PLACEHOLDER_0__ the sides through the vertex.

2.Show that the line connecting point __A_TAG_PLACEHOLDER_0__Ato where the lines areCFandBGintersect is an axis of symmetry for the shape.

3.If two isosceles triangles share the same base and are either on the same side or opposite sides of it, the line that connects their vertices serves as a line of symmetry for the shape formed by them.

4.Show how to prove this proposition by assuming that every angle has a __A_TAG_PLACEHOLDER_0___. bisector.

5.Each diagonal of a diamond shape is a line of symmetry for that diamond.

6.If you place three points on the sides of an equilateral triangle, with one point on each side and evenly spaced from the corners, the lines that connect those points will form a new equilateral triangle.

Dem.—If AB and AC aren't equal, one has to be greater than the other. Let’s say AB is the greater one, and that the part BD is equal to AC. Connect CD (Post. i.). Then the two triangles DBC and ACB have BD equal to AC, and BC is common to both. Thus, the two sides DB and BC in one triangle are equal to the two sides AC and CB in the other; and the angle DBC in one is equal to the angle ACB in the other (hyp). Therefore [i.v.] the triangle DBC is equal to the triangle ACB—the smaller to the larger, which is absurd; therefore AC,ABare not unequal, which means they are equal.

1.What is the hypothesis in this statement?

2.What proposition does this contradict?

3.What is the opposite of this statement?

What's the opposite of Prop.?

5.What does an indirect proof mean?

6.How does Euclid usually prove converse propositions?

7.What wrong assumption is made in the demonstration?

8.What does this assumption lead to?

If two triangles(ACB,ADB) share the same foundation(AB) and are on the same side of it with one pair of adjacent sides(ACAD) equal to one another, the other pair of neighboring sides(Before Christ,BD) must be unfair.

Dem.—1. Let the vertex of each triangle be without the other. Connect CD. Then, since AD is equal to AC (hypotenuse), the triangle ACD is isosceles; therefore [v.] the angle ACD is equal to the angle ADC; but ADC is greater than BDC (Axiom ix.); therefore ACD is greater than BDC: moreover, BCD is much greater than BDC. Now if the side BD were equal to BC, the angle BCD would also be equal to BDC [v.]; but it has been proved to be greater. Therefore BD is not equal to BC.

2. Let the vertex of one triangle ADB be inside the other triangle ACB. Extend the sides AC and AD to points E and F. Since AC is equal to AD (the hypotenuse), triangle ACD is isosceles, and thus the external angles ECD and FDC on the other side of the base CDs are equal. However, ECD is greater than BCD (Axiom i.x.). Therefore, FDC is greater than BCD: even more so, BDC is greater than BCD. But if BC were equal to BD, then the angle BDC would be equal to BCD [v.]; therefore, therefore BC can't be equivalent to BD.

3. If the vertex D of the second triangle falls on the line BC, it is clear that BC and BD are not equal.

Not provided.What is the purpose of Prop? vi.?Ans.It acts as a lemma for Prop. viii.

2.In the demo of Prop. vi.the contrapositive of Prop. v.happens; display where.

3.Show that two circles can only intersect at one point on the same side of the line. connecting their centers, which means that two circles cannot intersect at more than two points.

If two triangles(ABC,DEF) have two sides to it(AB,AC) of one that are equal to two sides(DE,DF) of the other, and also hold the foundation(BC) of one equal to the headquarters(EF) of each other; then the two triangles will be equal, and their angles will one will be equal to the angles of the other—specifically, those angles will be equal to which the equal sides face.

Dem.—Let's apply triangle ABC to DEF, so that point B lines up with E, and line BC lines up with line EF; since BC is equal to EF, point C will line up with F. Now, if vertex A is on the same side of EF as vertex D, then point A must line up with D; if not, let it take a different position G; then we have EG equal to BA, and BA is equal to ED (hyp.). Therefore (Axiom i.) EG is equal to ED: similarly, FG is equal to FD, and this is impossible [vii.]. Thus, point A must coincide with D, which means triangle ABC is identical to triangle DEF; Therefore, the three angles of one are respectively equal to the three angles of the __A_TAG_PLACEHOLDER_0__. other—specifically,AtoD,BtoE, andCtoF, and the two triangles are equal.

This Proposition is the opposite of i.v., and it represents the second scenario of triangle congruence in the Elements.

Philo's Argument.—Let's use the same bases as shown in the previous proof, but position the vertices on opposite sides; letBGCrepresent the stance taken byEDF. ConnectAGSince BG=BA, the angleBAGI'm sorry, but there doesn’t seem to be any text for me to modernize. Please provide the phrases you'd like me to work on.BGA. Similarly, the angleCAGUnderstood. Please provide the text for modernization.CGA. So, the whole angle BACUnderstood. Please provide the text you would like me to modernize.BGC; butBGCUnderstood! Please provide the text you would like me to modernize.EDF, soBAC=EDF.

Sun.—In AB choose any point D, and cut off [iii] AE equal to AD. Connect DE (Post. I.), and on the side opposite to A, draw the equilateral triangle DEF[i.] Connect AF. AFdivides the given angleBAC.

Dem.—The triangles DAF and EAF have side AD equal to AE (constant) and share side AF; therefore, the two sides DA and AF are equal to EA and AF respectively, and the base DF equals the base EF, since they are the sides of an equilateral triangle (Def. xxi.). Therefore [vii.] angle DAF is equal to angle EAF; so, angleBAC is divided by line AF.

Cor.—The line AFis a line of symmetry for the figure.

1.Why does Euclid describe the equilateral triangle on the opposite side?A?

2.In what situation would the construction fail if the equilateral triangle was drawn on the __A_TAG_PLACEHOLDER_0__? other side ofDE?

1.Prove this statement without using Prop. viii.

Prove that AF is perpendicular to DE.

3.Show that any point onAFis the same distance from the pointsDandE.

4.Show that any point onAFis equidistant from the linesABandAC.

Sol.—On AB, draw an equilateral triangle ACB[i.]. Split the angle ACB in half using the line CD[i x.], which meets AB at D, then AB is divided at D.

Dems.—The two triangles ACD and BCD have side AC equal to Before Christ, which are the sides of an equilateral triangle, and CD is common. Therefore, the two sides AC and CD in one triangle are equal to the two sides BC and CD in the other; and the angle ACD is equal to the angle BCD (const.). Thus, the base AD is equal to the base DB [i.v.]. So AB is divided at D.

1.Show how to split a finite straight line in half by drawing two circles.

2.Any point that is equidistant from the pointsAandBis located on the lineCD.

From a certain point(C) on a specific straight line(AB) to draw a straight line that is at a right angle to the given line.

Sun.—In AC, pick any point D, and make CE equal to CD [iii.]. From DE, draw an equilateral triangle DFE [i.]. Connect CF. Then, CF will be at a 90-degree angle to AB.

Dem.—The two triangles DCF and ECF have CD equal to CE (constant) and share CF; therefore, the two sides CD and CF in one triangle are respectively equal to the two sides CE and CF in the other triangle, and the base DF is equal to the base EF, being the sides of an equilateral triangle (Def. xxi.); therefore, [viii.] the angle DCE is equal to the angle ECF, and they are adjacent angles. Thus (Def. xiii.) each of them is a right angle, andCF is at a right angle toAB at this point C.

It seems there was no text provided for me to modernize. Please provide a short phrase, and I'll be happy to assist!The diagonals of a rhombus cross each other at right angles.

2. Prove Proposition. xi. without using Proposition. viii.

3.Draw a line that is perpendicular to a given line at one of its ends without extending the line.

4.Find a location on a specific line that is equidistant from the two given points.

5.Locate a point on a line so that, when it connects to two designated points on opposite sides of the line, the angle formed by those lines is bisected by the line itself.

6.Find a point that is equidistant from three given points.

Sol.—Choose any point D on the opposite side of AB, and draw (Post. iii.) a circle, with C as the center and CD as the radius, intersecting AB at the points F and G. Bisect FG at H [x.]. Connect CH (Post. i.). CH will be at right angles to AB.

Dem.—Join CF and CG. Then the two triangles FHC and GHC have FH equal to GH (constant), and HC is common; the base CF is equal to the base CG, as they are both radii of the circle FDG (Def. xxxii.). Therefore, the angle CHF is equal to the angle CHG [viii.], and since they are adjacent angles, they are right angles (Def. xiii.). So CH is at a right angle toAB.

1.Demonstrate that the circle cannot intersect.ABat over two points.

2.If one angle of a triangle is equal to the sum of the other two angles, the triangle can be divided. into two isosceles triangles, and the base is twice the length from its midpoint. to the opposite corner.

The adjacent angles(ABC,ABD) made by one straight line(AB) standing on another one (CD) are either both right angles, or their sum equals two right angles. angles.

Dem.—If AB is perpendicular to CD, as shown in fig. 1, the angles ABC and ABD are right angles. If they aren't, draw BE so that it is perpendicular to CDs [xi.]. Now, the angle CBA is equal to the sum of the two angles CBE and EBA (Def. xi.). Therefore, if you add the angle ABD, the total of the angles Can't be arsed and ABD is equal to the sum of the three angles CBE, EBA, and ABD. Similarly, the sum of the angles CBE and EBD is equal to the sum of the three angles CBE, EBA, and ABD. Things that are equal to the same thing are equal to each other. Therefore, the sum of the angles CBA and ABD is equal to the sum of the angles CBE and EBD; but CBE and EBD are right angles; So, the total of the anglesCBA,ABDis two correct angles.

Or instead:Let's name the angleEBAasθ; then obviously

Cor. 1.—The total of two supplementary angles is equal to 180 degrees.

Cor. 2.—Two straight lines can't share a common segment.

Cor. 3.—The bisector of any angle divides the corresponding re-entrant angle into two equal parts.

Cor. 4.—The bisectors of two supplementary angles are perpendicular to each other.

Cor. 5.—The angle EBA is half the difference between the angles CBA and ABD.

If at some point(B) in a straight line(BA) two more straight lines(CB,BD) on opposite sides create adjacent angles(CBA,ABD) that total two, right? When these two straight lines are joined, they create one continuous line.

Dem.—If BD is not the continuation of CB, then let BE be its continuation. Now, since CBE is a straight line, and BA stands on it, the sum of the angles CBA and ABE is two right angles (xiii.); and the sum of the angles CBA and ABD is two right angles (hyp.); therefore the sum of the angles CBA and ABE equals the sum of the angles CBA and ABD. If we remove the angle Can't be bothered, which is common, we find that the angle ABE equals the angle ABD—that is, a part equals the whole—which is absurd. So, BD must be on the same straight line as CB.

Dem.—Since the line AE lies on CD, the total of the angles CEA and AED is two right angles [xiii.]; and since the line CE lies on AB, the total of the angles BEC and CEA is two right angles; therefore, the total of the angles CEA and AED equals the total of the angles BEC and CEA. Eliminate the angle CEA, which is common, and we find that the angleAEDequalsBEC. Similarly, the perspectiveCEAis equal toDEB.

The previous proof can be summarized by stating that opposite angles are equal because they share a common supplement.

1.What is the issue needed in Euclid’s proof of Prop? xiii.?

2.Which theorem?Sure! Please provide the text you would like me to modernize.No theorem, just the rules.

3.If two lines intersect, how many pairs of supplementary angles are formed?

4. How does Prop. XIV relate to Prop. XIII?

5.Which three lines in Prop? x i v.are they happening at the same time?

6.What attention is required when stating Prop. __A_TAG_PLACEHOLDER_0__? xiv.?

7.State the opposite of Prop. xvProve it.

8.What is the topic of Props? xiii., xiv., xv.?Understood. Please provide the text you want me to modernize.Angles at a point.

PROP. XVI.—Theorem.

If either side(Before Christ) triangle(ABC) is extended, the outside angle(ACD) is greater than either of the interiors non-adjacent angles.

Dem.—Divide ACat E[x.]. Connect Be.(Post. i.). Extend it, and from the extended part cut off EF equal to Be[iii]. Connect CF. Now because EC is equal to EA (constant), and EF is equal to EB, the triangles CEF and AEB have sides CE, EF in one equal to the sides AE, EB in the other; and the angle CEF is equal to AEB[xv.]. Therefore [iv.] the angle ECF is equal to EAB; but the angle ACD is greater than ECF; therefore the angle ACD is greater than EAB.

Similarly, it can be shown that if side AC is extended, the exterior angle BCG is greater than angle ABC; however, BCG is equal to ACD [xv.]. Therefore, ACD is greater than ABC. Thus, ACD is greater than either of the inside non-adjacent angles A or B of the triangle ABC.

Cor.Please provide the short text you would like me to modernize.—The total of the three interior angles of triangle BCF is the same as the total of the three interior angles of triangle ABC.

Cor.2.—The area of BCF is the same as the area of ABC.

Cor.3.—The lines BAand CF, if extended, cannot meet at any finite distance. If they did meet at some finite point X, then the triangle CAXwould have an exterior angle BACequal to the interior angle ACX.

Dem.—Create BCto D; then the exterior angle ACDis larger than ABC [xvi.]: add the angle ACB to each, and we find that the total of the angles ACD, ACB is greater than the total of the angles ABC, ACB; but the total of the angles ACD, ACBis two right angles [xiii.]. So, the total of the anglesABC,ACBis fewer more than two right angles.

In the same way, we can demonstrate that the sum of the angles A, B, or the angles A, C, is less than two right angles.

Correct.1.—Every triangle has to have at least two acute angles.

Cor.2.—If two angles of a triangle are not equal, the smaller one must be acute.

Prove Prop. xvi. without producing a side.

In any triangle(ABC) if one side(AC) is longer than the other side(AB), then the the angle opposite the longer side is greater than the angle opposite the shorter side.

Dem.—From AC cut off AD equal to AB[iii]. Connect BD (Post. i.). Now since AB is equal to AD, triangle ABD is isosceles; therefore [v.] angle ADB is equal to angle ABD; but angle ADB is greater than angle ACB[xvi.]; therefore ABD is greater than ACB. Also, angleABCis much greater than angle ACB.

Or alternatively:From Aas the center, with the shorter side AB as the radius, draw the circle Bed, intersecting BC at E. Connect AE. Now since AB is equal to AE, the angle AEB is equal to ABE; but AEB is greater than ACB(xvi.); accordinglyABEis more than ACB.

1.If in the second method the circle crosses the lineCBextended throughB, prove it Proposal.

2.This idea can be shown by showcasing the smaller side.

3.If two opposite sides of a quadrilateral are the longest and the shortest, the angles __A_TAG_PLACEHOLDER_0__ The angles next to the shortest side are greater than the angles opposite them.

4.In any triangle, the perpendicular line drawn from the vertex opposite the longest side will be inside the triangle.

If one angle(B) of a triangle(ABC) is larger than another angle(C), then the side (AC) The side opposite the larger angle is longer than the other side.(AB) across from the smaller angle.

Dem.—If ACis not greater than AB, it must either be equal to it or less than it. Let's look at each case:—

1. If ACwere the same as AB, the triangle ACBwould be isosceles, which would mean the angle Bwould equal C[v]; but that's not the case by assumption; therefore ABis not the same as AC.

2. If AC were shorter than AB, then angle B would be smaller than angle C[xviii.]; but that’s not the case according to our assumption; so AC is not less than AB; and since AC is neither equal to AB nor shorter than it, it has to be better.