Book 1 of The Elements

Book 1 of the Elements

Euclid’s axioms: http://aleph0.clarku.edu/~djoyce/java/elements/bookI/bookI.html • Postulate 1. “To draw a straight line from any point to any point.” • Postulate 2. “To produce a finite straight line continuously in a straight line.” • Postulate 3. “To describe a circle with any center and radius.” • Postulate 4. “That all right angles equal one another.” • Postulate 5. “That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.”

Proposition 1

- Given points A and B create circles centered at A passing through B and centered at B passing through A(Postulate 3) - Make the point C where the circles intersect - The two circles with one centered at A and one centered at B have the same radii of length AB - Note AC, CB, and BA are all radii of equal circles - With all the sides of the triangle being of the same length, triangle ABC is an equilateral triangle

Proposition 2

-Given segment AB and point P construct segment between B and P(Postulate 2) -Construct equilateral triangle BDC(Proposition 1) -Extend segments CD and CB to lines(Postulate 2) -BA and BE are equal segments as they are radii of the same circle -Construct a circle centered at D passing through E, and mark the point of intersection, F, between this circle and the line passing through DF. -FD and DE have the same length, because they are radii of the same circle, centered at D. -CP and CB are congruent because they are sides of equilateral triangle ABC. -The length FD minus the length of CD is equal to the length of DE minus the length of BD because equals subtracted from equals are equal(common notion). -Thus, DF and AB are congruent

Proposition 3

-Given two segments AB and B_1C_1 where one is greater than the other. -WLOG, let segment AB be the longer one of the two. -Place BC at the end of AB. -Draw a circle centered at B passing through C and label the intersection point D. -Because both segments DB and BC are radii of the same circle they are equal. -These segments are also equal to the original segment B_1C_1 -Therefore, given the two straight lines AB and B_1C_1, AD has been cut off from AB the greater equal to the less.

Proposition 4(SAS)

-Given two triangles ABC and DEF with two equal sides and an equal angle between the two sides. -The third side of both triangles must be equal as two equal sides with an between can only create one triangle. -Because all the side lengths are equal, all of the angles are equal respectively. -Thus, the triangles share all sides and angles.

Proposition 16

In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Step Justification Side AE = Side EC Proposition 10 Side BE Postulate 1 Side BF Postulate 2 Side BE= Side EG Proposition 3 Side CG Postulate 1 Angle AEB= Angle GEC Proposition 15 Triangle AEB = Triangle CEG Proposition 4 Angle BAE= Angle GCE Thus, Angle ACD is greater than Angle ACB, Angle CAB, and Angle ABC. Also similarly, we can extend AC and find anther external angle that is greater than angle BAC. By Proposition 15, the two external angles are congruent which means the first angle is also greater than ABC.

Proposition 17

In any triangle the sum of any two angles is less than two right angles. Steps Justification Produce Segment CD Postulate 2 Angle ACD

greater than Angle ABC Proposition 16 ACD+ACB greater than ABC+ACD. ACD+ACB= 2 Right Angles 2 Right Angles greater than ABC+ACD Thus, the sum of the angles ABC and ACD are less than 2 Right Angles. Also similarly, this can be done for any of the angles in any triangle.

Proposition 5

-AF equals AG, and AB equals AC, so FA and AG are equal. The two sides GA and FA are equal because equals subtracted from equals are equal. -They contain a common angle, the angle GAF. -So triangle AFC equals the triangle AGB. -FC is equal to GB so the two sides BF and FC equal the two sides CG and GB respectively -The angle BFC equals the angle CGB and share side BC -Triangle BFC also equals the triangle CGB -Therefore the angle FBC equals the angle GCB, and the angle BCF equals the angle CBG. -Angle ABG was proved equal to the angle ACF, and in these the angle CBG equals the angle BCF, the remaining angle ABC equals the remaining angle ACB, and they are at the base of the triangle ABC. -The angle FBC was also proved equal to the angle GCB, and they are under the base. -In an isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.

Proposition 6

If a triangle has two equal angles, then the sides opposite are also equal. - Given a triangle with two equal base angles, the two corresponding sides must be equal. - We will prove this by contradiction. - If the sides are not equal, one side is longer than the other. - Cut the shorter side length from the longer one. - In the example above, DB equals AC, and BC is common, therefore the two sides DB and BC equal the two sides AC and CB respectively, and the angle DBC equals the angle ACB. - Therefore the base DC equals the base AB, and the triangle DBC equals the triangle ACB, the less equals the greater.(Contradiction) - Thus AB is not unequal to AC, it therefore equals it.

Proposition 7

- Assume toward a contradiction that there exists a point D where AD is equal in length to AC and BD is equal to BC. - Connect CD and DB to create triangle CDB and connect AD to create triangle ABD. - By definition both triangles ACD and ABD are isosceles(under our assumption) and have equal base angles. - However, looking at the construction we can see angle DCA is larger than angle DCB. - Similarly for angles CDA and CDB. - But since angle CDA is equal to DCA, and angle DCB is equal to angle CDB we have CDA both larger and smaller than DBC at the same time.(Contradiction) - Thus, given a triangle ABC, there is a unique point C where the sides of a triangle AC and BC meet.

Proposition 8

- (SSS) - If BC=EF and ED=BA and DF=AC, then the two triangles are equal in all aspects. - We will prove this using Propositions 2 and 7. - Using Proposition 2, we will copy each side to a corresponding end point on the side and rotate the copy so that it overlaps and fully intersects the other side. - By Proposition 7, the point of two sides of a triangle can only meet at one unique point. - This allows us to conclude that since all the endpoints of the lines are congruent, then the angles must also be congruent making these triangles equal in all aspects.

Proposition 9

- To bisect a given angle - Points B and C are constructed to be equidistance from point A as they are radii from the same circle. - Points B and C are also equidistance from point D since they are sides of an equilateral triangle. - By Proposition 8, the two triangles are congruent because they have three sides that are equal. - This also means the corresponding angles are equal. - Thus, the lines segment AD is a bisection of the angle CAB.

Proposition 10

- To bisect a line segment - We will prove this proposition by using Proposition 9 and 4. - Construct an equilateral triangle with side length AB and label C. - Use Proposition 9 to bisect angle ACB. - AC equals BC and angle ACD equals angle BCD. - Since the two triangles have a sides equal, an angle equal, and share a common side, the triangles are congruent by Proposition 4. - Thus, segment AD is equal to BD and we have bisected the segment AB.

Proposition 11

- To draw a straight line at right angles to a given straight line from a given point on it. - Construct points E and D equidistance from C on segment AB. - Construct a equilateral triangle with distance ED and label the third point F and connect F to point C. - Line DC equals line CE since they are of radii of the same circle. - FE and FD are equal because they are sides of an equilateral triangle. - Therefore we know triangles EFC and EFD are congruent triangles. - Thus, angles FCD and FCE are equal. - By the definition of a right angle, the two angles FCE and FCD are both right.

Proposition 12

- To draw a straight line perpendicular to a given infinite straight line from a given point not on it. - Construct points E and F on the given line by constructing a circle centered at C passing through G. - Lines CE and CF are equal because they are radii of the same circle centered at the given point C. - Then using proposition 10, bisect the line EF to get point H and segments EH and HF are equal. - Since the two triangles EHD and DHF have three congruent sides, they are equal triangles with equal angles. - Since the triangles are equal and EF is a line, by definition the angles are right angles and CG is perpendicular to EF.

Pythagorean Theorem

Pythagorean Theorem Proof

- Above is the image of reference for this proof. - We will be trying to prove that a^2 + b^2 = c^2. - Consider the image on the right. - Assume all the side lengths are touching and all right angles are right angles. - The area of the entire square is the 4 triangles plus the square in the middle. - The area of a triangle is .5(ab) and the square is c^2 and the big square is (a+b)^2 - Using the labels, we can write 4(.5(ab)+c^2 = (a+b)^2 - Simplifying, 2ab + c^2 = a^2 + 2ab + b^2 - Simplifying again by subtracting 2ab from each side we get, c^2 = a^2 + b^2. - Now comparing the two images... (not a real proof) - Both big squares have an area of (a+b)^2. - By reformatting the shapes within the large square and lengths along the sides, we can form 4 triangles with length b and height a next to each other. - The left over area is made up of two squares with area a^2 and b^2. - We can see that if you take away the yellow area(4 triangles), we get the green area left over which is equal in both images. - In one image the area is c^2 and in the other it is a^2 + b^2.

Book 1 of The Elements