parallel lines

Book 1 – Proposition 31

Through a given point to draw a straight line parallel to a given straight line.
[Desmos graphs can be found here]

Let A be the given point and BC be the given line. Our goal is to draw a line that is parallel to BC and passes through A.

Take a point D at random on BC, and connect AD.

On the straight line AD, at the point A, let the angle DAE be constructed such that it is equal to angle ADC by Proposition 23. (On a straight line and a point on it to construct an angle equal to a given angle.)
Let the straight line AF be produced in a straight line with EA.

The line AD falls on the two straight lines AB and EF,
The alternate angles DAE and ADC are equal to one another.
By Proposition 27 (If a straight line falls on two straight lines such that the alternate angles are equal, then the two lines are parallel to one another), the line EF is parallel to the line BC.

Therefore we have constructed a line that passes through the given point A and is parallel to the given line BC.

We will use this proposition to prove that the three angles in a traingle are equal to two right angles – George

Book 1 – Proposition 30

Straight lines parallel to the same straight line are also parallel to one another.
[Desmos graphs can be found here]

If the two lines AB and CD are both parallel to another line EF, we will prove that AB and CD are parallel to each other.

Let the straight line GK fall upon the three lines as shown.

By Proposition 29 (alternate angles are equal), since AB and EF are parallel angle AGK is equal to the angle GHF.

Since EF is parallel to CD, Proposition 29 tells us that angle GHF is equal to angle GKD. (Exterior angle is equal to the interior and opposite angle.)

Therefore angle AGK is equal to angle GKD. (AGK = GHF = GKD)

Since angles AGK and GKD are alternate, Proposition 27 tells us that the lines AB and CD are parallel to one another. (If a straight line falls on two straight lines and the alternate angles are equal, then the two straight lines are parallel.)

Q.E.D.

The next proposition involves the construction of a line through a given point that is parallel to a given line – George

Book 1 – Proposition 29

A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the opposite and interior angle, and the interior angles on the same side equal to two right angles.

Let EF fall on the parallel straight lines AB and CD.

Part 1: We will prove that the alternate angles AGH and GHD are equal.

If the two angles are not equal, then one must be greater. Let AGH be the greater angle.
Add angle BGH to both, and angles AGH and BGH are greater than angles GHD and BGH. But the angles AGH and BGH are equal to two right angles by Proposition 13 (If a straight line falls on a straight line, it makes two angles that are equal to two right angles.)
Therefore the two angles GHD and BGH are less than two right angles. Lines that produce indefinitely from angles less than two right angles meet (this is one of Euclid’s Postulates), meaning that AB and CD will meet.
But AB and CD cannot meet because they are parallel, and this contradiction implies that angle AGH cannot be greater than angle GHD.

Therefore the alternate angles AGH and GHD must be equal.

Part 2: We will prove that the exterior angle EGB is equal to the interior and opposite angle GHD.

In Part 1 we proved that angle AGH is equal to angle GHD.
By Proposition 15 (vertical angles are equal to each other) angle AGH is equal to angle EGB.

Therefore the exterior angle EGB is equal to the interior and opposite angle GHD.

Part 3: We will prove that the interior angles on the same side (BGH and GHD) are equal to two right angles.

In Part 2 we proved that angle EGB was equal to angle GHD. Add angle BGH to each, and we have that the two angles EGB and BGH are equal to the two angles GHD and BGH.

Again using Proposition 13 (a straight line that falls on a straight line produces two angles that are equal to two right angles), the two angles EGB and BGH are equal to two right angles. That implies that the two angles BGH and GHD are equal to two right angles.

Therefore the interior angles on the same side are equal to two right angles.

Q.E.D.

More fun with parallel lines in the next Proposition – George

Book 1 – Proposition 28

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

Let the straight line EF fall on the two straight lines AB and CD, intersecting AB at a point G and intersecting CD at a point H, is such a way that exterior angle EGB is equal to the interior and opposite angle GHD or the interior angles on the same side (BGH and GHD) are equal to two right angles.

Starting with angle EGB being equal to angle GHD we will prove that AB is parallel to CD.

Angle EGB is equal to angle GHD by hypothesis.
Angle EGB is also equal to angle AGH by Proposition 15 (vertical angles are equal)

So, angle GHD is equal to angle AGH. Since these two angles are alternate angles, AB and CD are parallel by Proposition 27. (If a line falling on two straight lines produces equal alternate angles then the two lines are parallel.)

Now we will start with angles BGH and GHD being equal to two right angles, and will once again prove that AB and CD are parallel.

Angles AGH and BGH are also equal to two right angles by Proposition 13. (A straight line set up on a straight line will make angles equal to two right angles.)

So, the angles AGH and BGH together are equal to the angles BGH and GHD together. Subtracting angle BGH from each, the remaining angle AGH is equal to the remaining angle GHD.

Since angles AGH and GHD are alternate angles, Proposition 27 again tells us that AB and CD are parallel.

Q.E.D.

Book 1 – Proposition 27

If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.
[Desmos graphs can be found here]

Let line EF fall on the straight lines AB and CD with the alternate angles AEF and EFD equal to one another.

We will prove that AB is parallel to CD by showing that they cannot intersect. Suppose that they did intersect at a point G in the direction of B and D.

We will look at triangle GEF. By Proposition 16 (In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles) the exterior angle AEF must be greater than the interior angle EFG. But, by hypothesis, those two angles are equal which produces a contradiction.

Therefore AB and CD do not meet at a point in the direction of B and D. In a similar fashion it can be shown that they do not intersect at a point in the direction of C and D.

Thus, AB is parallel to CD.

Q.E.D.

This proposition will be used in the proof of Proposition 28 which states that a line that intersects two parallel lines produces alternate angles that are equal to one another.