desmos – Page 4

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

Book 1 – Proposition 7

Proposition 7

Given two straight lines constructed on a straight line [from its extremities] and meeting in a point, there cannot be constructed on the same line [from its extremities], and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

Desmos.com link to graphs

The strategy for this proof is to perform a proof by contradiction. We will assume that two other straight lines can be constructed in this fashion and will reach a contradiction.

Suppose two straight lines AC and CB are constructed on the straight line AB and meeting at the same point C as shown.

Basically, we have started with a line segment AB and added a point C to complete a triangle. Proposition states that we cannot add another point D on the same side of AB such that AC is equal to AD and BC is equal to BD. Let’s add a point D on the same side of AB as C is.

This will be a proof by contradiction. Assume that AC is equal to AD and BC is equal to BD.

Since AC is equal to AD, the angle ACD is equal to the angle ADC by Proposition 5.

Since angle DCB is smaller than angle ACD (because it is contained in that angle), we know that angle ADC is greater than angle DCB.

The following show that angle CDB is greater than angle DCB:

Angle CDB is greater than angle CDA (because angle CDA is contained in angle CDB).
Angle CDA is equal to angle ACD as stated previously.
Angle ACD is greater than angle DCB (because angle DCB is contained in angle ACD).

[CDB > CDA = ACD > DCB]

Now we shift our attention to triangle BCD. Since BC is equal to BD, Proposition 5 tells us that angle CDB is equal to angle DCB.

But we just showed that if our assumptions were true that angle CDB is greater than angle DCB, which is a contradiction. Therefore, there cannot be constructed on the same line [from its extremities], and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

Q.E.D.

That wraps up one week’s worth of these propositions. I hope you are finding it rewarding – George

Book 1 – Proposition 6

Proposition 6

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

(Note: If you’d like to click through the images in this proof, they are available in this graph on Desmos.com.)

In Proposition 5 we started with two equal sides and proved that the angles that were opposite of those sides were equal. In this proposition we will begin with two equal angles and prove that the two sides that are opposite those angles are equal. We will begin with the triangle ABC, and let angle ABC be equal to angle ACB.

This proof is an example of a proof by contradiction. We will assume that sides AB and AC are not equal to one another, and as we proceed we will run into a contradiction that will tell us that our initial assumption is false and that sides AB and AC must be equal to one another.

Assume that sides AB and AC are not equal to one another. In that case one side must be longer than the other. Suppose that side AB is longer than side AC. From the longer side AB let DB be cut off equal to AC. (Proposition 3)
Join the points C and D as shown.

We will now be comparing the triangles DCB and ACB.
We know that DB is equal to AC, and the two triangles have the side BC in common. Also, the angle DBC is equal to the angle ACB as those are the two equal angles from triangle ABC.

Since the two triangles have two pairs of equal sides with an equal angle between them, Proposition 4 tells us that triangle DCB is equal to triangle ACB. However triangle DCB is clearly less than triangle ACB, so we have reached a contradiction. Our assumption that AB is not equal to AC is absurd, and therefore the two sides AB and AC must be equal to each other.

Q.E.D.

Proof by contradiction is an often used form for proving propositions or theorems in mathematics. The U.S. court system uses this principle in its trials – we assume that the defendant in innocent and then search for evidence that contradicts this assumption. A statistical hypothesis test uses the same format – assuming that the null hypothesis is true prior to searching for evidence that it is false.

Note: I have decided to stop previewing propositions. I will cover one proposition per post – George