desmos – Page 3

Book 1 – Proposition 23

On a given straight line and a point on it to construct a rectilineal angle equal to a given rectilineal angle.
[Desmos graphs can be found here]

We begin with a line AB (let A be the point on it) and an angle DCE where D and E are selected at random.

The goal is to construct an angle on AB at point A that is equal to angle DCE.

Begin by joining DE. Construct a triangle AFG on AB such that AF, FG, and GA are equal to CD, DE, and EC respectively. (Proposition 22 shows how such a triangle can be constructed with sides equal to those 3 line segments.)

The two sides DC and CE are equal to the two sides FA and AG, and the base DE is equal to the base FG. By Proposition 8 (Two triangles with two sides equal to two sides respectively and equal bases have equal angles between the two sides) angles DCE and FAG are equal.

Thus we have constructed an angle on AB that is equal to the given angle.

Book 1 – Proposition 22

Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. (Proposition 20)
[Desmos step-by-step graphs can be found here]

Let three lines A, B, and C satisfy the condition that any two taken together are greater than the remaining one.

We will now construct a triangle whose sides are equal to these three lines.
Set out a straight line beginning at a point D that continues through point E.
Place a point F such that DF is equal to A, a second point G such that FG is equal to B, and a third point H such that GH is equal to C.
(Proposition 3 allows us to create a line segment of length equal to the length of a given line segment.)

Construct a circle centered at F with a radius equal to DF. This circle can be labeled as DKL.

Construct a second circle centered at G with a radius equal to GH. This circle can be labeled as HKL, where K and L are the two points of intersection between the two circles.

Let FK and GK be joined.

We will now show that triangle FGK has three sides that are equal to A, B, and C.

Since K is on the circle centered at F, FK is equal to the radius DF which is also equal to A. Therefore FK is equal to A.
Since K is also on the circle centered at G, GK is equal to the radius GH which is also equal to C. Therefore GK is equal to C.
FG was set up to be equal to B.

FK, FG, and GK are equal to the three straight lines A, B, and C.

Therefore a triangle with sides equal to the three lines has been constructed.

Does triangle FGL also satisfy the requirements? As an exercise, see if you can show that triangle FGL has sides that are the lengths of A, B, and C.

The next proposition is another construction, requiring us to construct an angle that is equal to a given angle. The construction in Proposition 22 plays a large role in that construction.

George

Book 1 – Proposition 21

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
[Desmos graphs can be found here]

Just a quick translation here – If you start with a triangle, and from its base construct a new triangle whose third vertex is located within the original triangle, then the two new sides will be less than the original two sides but the angle contained between them will be greater.

We will start with a triangle ABC, and construct a triangle BCD on the base BC such that D is within the triangle.

We will prove that the two sides BD and DC are less than BA and AC, and that the angle BDC is greater than the angle BAC.

Let BD be drawn through to a point E that lies on AC.

In triangle ABE, Proposition 20 (two sides of a triangle together are greater than the remaining side) tells us that sides AB and AE together are greater than BE.
If we add EC to each, we have that AB and AC (AE & EC) together are greater than BE and EC.

We now have that AB and AC greater than two other segments (BE and EC). We need to show that those two sides are also greater than BD and DC.

Looking to triangle CDE, Proposition 20 tells us that the two sides CE and ED together are greater than DC.
If we add BD to each we have that EC and BE (BD & ED) together are greater than DC and BD.

Therefore AB and AC together are greater than BD and DC.
[ AB & AC > BE & EC > BD & DC ]

Now we move on to prove the statement concerning the angles.
(Angle BDC is less than angle BAC)

We will begin by using Proposition 16 (exterior angle of a triangle is greater than either of the opposite interior angles) in triangle CDE.

Angle BDC (exterior angle) is greater than angle CED (an opposite interior angle). Angle CED is equal to angle CEB, so we have angle BDC as greater than angle CEB.

In triangle ABE, the exterior angle CEB is greater than the opposite interior angle BAE (which is equal to angle BAC).

Since angle BDC is greater than angle CEB, which is greater than angle BAC, therefore angle BDC is greater than angle BAC.

Q.E.D.

Book 1 – Proposition 20

In any triangle two sides taken together in any manner are greater than the remaining one.
[Desmos graphs available here]

Intuitively this should make sense, because if the two sides were shorter than the third they would not even cover the third segment laid end to end, making it impossible to form a triangle.

For a triangle ABC, we will prove the sides AB and AC are greater than side BC. A similar strategy can be used to show that AB and BC are greater than AC, and that AC and BC are greater than AB.

Extend the segment BA through a point D such that the segment AD has the same length as segment AC. (Proposition 2 allows us to construct a line segment of the same length as AC.)
Let DC be joined.

Look at triangle ACD. Since AD is equal to AC, triangle ACD is an isosceles triangle. By Proposition 5, angle ADC is equal to angle ACD. (The base angles of an isosceles triangle are equal.)

Because angle BCD is greater than angle ACD, we know that angle BCD is greater than angle ADC. We also know that angles ADC and BDC are equal.

Looking at triangle BCD …
Since angle BCD is greater than angle BDC, side BD is greater than side BC by Proposition 19 (the greater angle is subtended by the greater side).
The side BD is equal to AD and AB.
Since AD was created equal to AC, side BD is equal to AC and AB.
Thus AC and AB together are greater than BC.
[ BD > BC … AD & AB together > BC … AC & AB > BC]

Q.E.D.

We showed that two arbitrarily selected sides together are greater than the remaining side, so our results apply to each pairing in the triangle.

Questions? Comments? You know what to do – George

Book 1 – Proposition 18

In any triangle the greater side subtends the greater angle.
[Desmos graphs can be found here]

Let ABC be a triangle with aide AC greater than AB.

We will prove that angle ABC is greater than angle ACB.

Since AC is greater than AB, put a point D on AC such that AD is equal to AB.

Angle ADB is an exterior angle of triangle BCD, so by Proposition 16 it is greater than the opposite interior angle DCB.
(Proposition 16 stated that the exterior angle is greater than either of the opposite interior angles.)

Angle ADB is equal to angle ABD, because AB and AD are two equal sides in a triangle.
So, angle ABD is greater than angle ACB (which is the same as angle DCB).
Since angle ABC is greater than angle ABD, angle ABC is also greater than angle ACB.
[ ABC > ABD = ADB > DCB = ACB ]

Q.E.D.

We just proved that the greater side subtends the greater angle.
In Proposition 19 we will prove that the greater angle is subtended by the greater side.

I appreciate any feedback or questions – George

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 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