January 2016 – Page 2

Book 1 – Proposition 17

In any triangle, two angles taken together in any manner are less than two right angles.
[Desmos graphs can be found here]

In this proof we will show that this holds true for two arbitrarily chosen angles, which can then be extended to any two angles in the triangle. The proof does use Proposition 16 (two opposite interior angles are each less than the exterior angle produced from the third angle) and Proposition 13 (a straight line set upon another straight line makes two angles that are equal to two right angles).

Let ABC be a triangle. We will start by showing that the two angles ABC and BCA are less than two right angles.

Let BC be produced to a point D.

By Proposition 16 angle ACD is greater than angle ABC, because the exterior angle (ACD) is greater than either of the two opposite interior angles (including angle ABC).

Add angle BCA to both.
Thus the two angles ACD and BCA are greater than the two angles ABC and BCA.
But the two angles ACD and BCA are equal to two right angles. (By Proposition 13)
Therefore the two angles ABC and BCA are less than two right angles.

In a similar fashion we can show that the two angles BCA and CAB are less than two right angles, as are the two angles CAB and ABC.

Q.E.D.

Please leave any questions or comments in the Comment section – George

Book 1 – Proposition 16

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior or opposite angles.
[Desmos graphs can be found here]

Start with triangle ABC, and let one side of BC be produced to point D.

We will prove that the exterior angle ACD is greater than either of the interior and opposite angles BAC and CBA.

By Proposition 10 (that allows us to bisect a line segment) let AC be bisected at E.
Let BE be joined and produced to a point F so that EF is equal to BE. (Proposition 3 allows us to construct a line segment equal to a given segment.)
Let FC be joined.

Finally, let the side AC be drawn through to a point G as shown.

We will now examine the two triangles ABE and CFE.

Sides AE and EC are equal, since E bisects AC.
Sides BE and EF are equal according to the way we selected point F.
Angles AEB and CEF are equal to each other because they are vertical angles.
(Proposition 15 states that vertical angles are equal.)

So, by Proposition 4 (“side-angle-side”), the side AB is equal to the side CF and the remaining corresponding angles are equal. In particular, angle BAE is equal to angle ECF.

Angle ECD is greater than angle ECF, and that implies that angle ACD is greater than angle BAE. (Angle ACD is equal to angle ECD, and angle BAE is equal to angle ECF.)

In a similar fashion, we can prove that angle ACD is greater than angle ABC.
(Bisect BC … angle BCG equals angle ACD (vertical angles) … construct triangles and show that angle BCG is greater than angle ABC.)

Q.E.D.

Later we will see that the three interior angles ABC, BCA, and ACB are equal to two right angles.
The adjacent angles ACB and ACD are also equal to two right angles.
That shows that the two angles ABC and BCA are equal to the angle ACD, and angle ACD must be greater than either of those two angles.

If you have feedback, I’d love to hear it – George

Book 1 – Proposition 15

Proposition 15

If two straight lines cut each other, they make vertical angles equal to one another.

This proposition appears a great deal in today’s geometry & trigonometry classes: Vertical angles are equal.

Let two lines AB and CD cut each other, intersecting at point E.
We will show that angle AEC is equal to angle BED, and that angle AED is equal to angle BEC.

Line AE lies on the straight line, so by Proposition 13 (When a straight line lies on a straight line, the two adjacent angles are equal to two right angles) the two angles AEC and AED are equal to two right angles.

Using the same reasoning, the two angles AED and BEDare equal to two right angles since the line DE lies on the straight line AB.

Since both pairs of angles are equal to two right angles, we know that the two angles AEC and AED are equal to the two angles AED and BED. Subtract the angle AED from each, and we find that angle AEC is equal to angle BED.

Can you prove that angle AED is equal to angle BEC? Try it first on your own, then click on the Proof Strategy tag below.

Please feel free to leave any questions/comments for me – George
Proof Strategy

Book 1 – Proposition 14

Proposition 14

If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.
[Desmos graphs can be found here]

We will start with a straight line AB, and from point B we will draw two lines BC and BD not lying on the same side. The adjacent angles ABC and ABD are equal to two right angles. We will show that BC and BD lie on a straight line.

Assume that BD is not on a straight line with BC. Let BE lie in a straight line with BC.

By Proposition 13 (If a straight line is set up on a straight line, it will make two right angles or two angles equal to two right angles), since AB lies on the straight line CBE the two angles ABC and ABE are equal to two right angles.

We began with ABC and ABD being equal to two right angles, so the two angles ABC and ABE are equal to the two angles ABC and ABD.

Subtract angle ABC from both, leaving angle ABE equal to angle ABD which is a contradiction. Therefore BE does not lie on a straight line with BC. There cannot be any other straight line except for BD that lies on a straight line with BC.

Therefore BC lies on a straight line as BD.

Q.E.D.

Book 1 – Proposition 13

Proposition 13

If a straight line set up on another straight line make angles, it will make either two right angles or angles equal to two right angles.
[Desmos graphs can be found here]

Let the straight line AB be set up on the straight line CD, making angles DBA and ABC as shown.

If angle DBA is equal to angle ABC, then those two adjacent angles are right angles.
Assume that they are not right angles. By Proposition 11 (which covers how to draw a straight line at right angles to another straight line ) we can draw the line segment BE that is at right angles to CD. So, angles DBE and EBC are two right angles.

We will now show that the two angles DBA and ABC are equal to two right angles.

Angle EBC is equal to the two angles EBA and ABC.
Adding angle DBE to each gives us that the two angles EBC and DBE are equal to the three angles EBA, ABC, and DBE. (Adding the same angle leaves equality.)

Angle DBA is equal to the two angles DBE and EBA.
Adding angle ABC to each gives us that the two angles DBA and ABC are equal to the three angles DBE, EBA, and ABC.

So, we have shown that
the three angles DBE, EBA, and ABC are equal to the two angles EBC and DBE
as well as
the three angles DBE, EBA, and ABC are equal to the two angles DBA and ABC.

Since things equal to the same thing are also equal to each other, the two angles EBC and DBE are equal to the two angles DBA and ABC.

Since the two angles EBC and DBE are equal to two right angles, the two angles DBA and ABC are also equal to two right angles.

So, we have proved that either DBA and ABC are two right angles, or they are equal to two right angles.

Q.E.D.

This proposition goes along with the common definition from algebra/geometry classes for supplemental angles. Two supplemental angles have a sum of 180°, which is also equal to the measure of two right angles.

Questions, comments, or feedback? You know the drill – please leave a comment! Thanks – George

Book 1 – Proposition 12

Proposition 12

To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.
[Link to graphs on Desmos]

Before we begin, here is the definition for right angles and perpendicular lines:
When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.To show that two lines are perpendicular we must show that the two lines form right angles.

Here is an infinite straight line AB and a point C that is not on it. The goal is to draw a line from point C that is perpendicular to AB.

Let point D be any point on the other side of AB from C, and draw a circle whose radius is equal to CD. I have labeled the circle as EFG, where points E and G are the points where the circle intersects the line AB.

Let the point H be the point at which the segment EG is bisected. (Proposition 10 covered how to find the point that bisected a line segment.)
Draw the segments from point C to the points G, H, and E.

We will now focus on triangles GCH and ECH as we try to show that the segment CH is perpendicular to AB.

GH is equal to HE, since H bisected EG.
CH is common to both triangles.
CG is equal to CE, because they are both on the circle with center C and radius CD.

So, by Proposition 8 (if two triangles have two equal sides and equal bases, then the angles between the equal sides are equal to each other, or in today’s introductory geometry classes: side-side-side implies equal angles), the two angles GHC and EHC are equal to each other.

Since the line segment CH set up on the straight line AB makes two angles GHC and EHC that are adjacent angles and equal to each other, both angles GHC and EHC are right angles.

By definition, the line segment CH is perpendicular to the line AB.

If you like what you see, or if you have suggestions for improvement, I would love to hear from you. Please leave any feedback as a comment – George

Book 1 – Proposition 11

Proposition 11

To draw a straight line at right angles to a given straight line from a given point on it.

[Desmos graphs are located here]

We will start with a line segment AB, and let C be the given point on it.

We will draw a line from point C at right angles from the segment AB.
We begin by plotting a point D on AC.
Let E be the point on CB such that CE is equal to CD. (Proposition 3 states that we can cut off a line segment that is equal to CD.)

On DE we can construct the equilateral triangle FDE (Proposition 1 shows the construction that shows an equilateral triangle can be built on any line segment), and join points F and C as shown.

We will now show that FC forms right angles with AB at point C by looking at triangles DCF and ECF.

DC is equal to EC by the way we chose point E.
CF is common to both triangles.
DF is equal to FE because they are both sides of the equilateral triangle DFE.

Therefore by Proposition 8 (if two triangles have three equal sides then the angles between two equal sides are equal), the angles DCF and ECF are equal to each other.

One of Euclid’s definitions will help us here: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the angles is right.

The two adjacent angles DCF and ECF are formed when a straight line (CF) is set up on a straight line (AB) and are equal to each other, so each of the angles is right.

This proposition will be used in the next proposition, a construction of drawing a line through a point that is perpendicular to a given straight line.

Questions, comments, or feedback? Please leave a comment below – George

Book 1 – Proposition 10

Proposition 10

To bisect a given straight line.

[Desmos.com graphs for this proposition]

Suppose we start with a line segment AB.

Construct the equilateral triangle ABC on it. (Proposition 1 covers how to construct an equilateral triangle om a line segment.)

Let angle ACB be bisected by the line segment CD, where the point D lies on the line segment AB. (Proposition 9 covers how to bisect a rectilinear angle.)

We will now compare the two triangles ACD and BCD.

AC is equal to BC because they are two sides of the equilateral triangle ABC.
Side CD is common to both triangles.
Angle ACD is equal to angle BCD, since CD bisects the angle ACB.

Since the two triangles have two equal sides, with equal angles between those equal sides, then the two bases AD and BD must be equal. (Proposition 4 states that if two triangles have equal side-angle-side, then the third sides must also be equal.)

Therefore AB has been bisected at D.

If you have any questions or feedback I’d love to hear it. Please leave a comment – George

Book 1 – Proposition 9

Proposition 9

To bisect a given rectilinear angle.

[Desmos graphs for this proposition]

This proposition is a construction that shows how to bisect an angle. We begin with the angle BAC.

Choose a point D at random on AB and cut off a segment AE from AC that is equal to AD. (Proposition 3)
Join DE.

Construct the equilateral triangle DEF be constructed using a point F on the opposite side of DE from point A . (Proposition 1)

Let AF be joined.

Now we will show that AF bisects the angle BAC by using the triangles DAF and EAF.

AD is equal to AE, since AE was cut off AC to equal AD.
AF is common to both triangles.
DF is equal to EF, since they are two of the sides of an equilateral triangle.

The conditions for Proposition 8 have been met, so angles DAF and EAF are equal to each other. Therefore the angle BAC has been bisected.

Now that we have learned how to bisect a rectilinear angle, we will learn how to bisect a given straight line (segment) in the next proposition.
As always, questions and feedback is welcome – George

Book 1 – Proposition 8

Proposition 8

If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

[Desmos graphs can be found here.]

Let ABC and DEF be two triangles such that AB is equal to DE and AC is equal to DF, and that the base BC is equal to the base EF.

We will prove that angle BAC is equal to angle EDF.

If triangle ABC is applied to triangle DEF with the point B placed on the point E and the straight line BC on the straight line EF, then the point C will also coincide with the point F because BC is equal to EF.
BA and AC will also coincide with ED and DF; if they did not then they would fall beside them and coincide with EG and GF for some other point G.

If that were true then given two straight lines (ED and DF) constructed on a straight line EF [from its extremities] and meeting at a point, there has been constructed on the same straight line [from its extremities] and on the same side of it two other straight lines (EG and GF) meeting at another point and equal to ED and DF. Proposition 7 tells us that this construction is not possible.

Therefore BA and AC must coincide with ED and DF. So, the angle BAC will coincide with the angle EDF. Thus the two angles BAC and EDF are equal to one another.

Q.E.D.

In this proof if triangle ABC coincided with triangle DEF, then the two angles BAV and EDF would be equal. So instead we assumed that the triangle ABC did not coincide with triangle DEF and showed that this was not possible.

The next four propositions will consist of constructions. I’d love to hear your feedback on Book 1 Proposition 8, or on anything about this blog – George