February 2016 – Page 2

Book 1 – Proposition 35

Parallelograms that are on the same base and in the same parallels are equal to one another.

We will start with two parallelograms on the same base BC and in the same parallels BC and AF.

We will prove that the parallelograms ABCD and EBCF shown below are equal.

Here are both parallelograms on the same graph.

By Proposition 34 (in a parallelogram opposite sides and opposite angles are equal) we know that AD is equal to BC as they are opposite sides in the parallalogram ABCD.
Looking to parallelogram EBCF, Proposition 34 tells us that EF is also equal to BC.
Therefore AD is equal to EF.Adding DE to each we find that AE (AD & DE) is equal to DF (DE & EF).

Again using Proposition 34, AB is equal to DC.

So EA is equal to FD, AB is equal to DC, and the angles contained (FDC and EAB) are equal by Proposition 29 (if a straight line falls on two parallel lines, the exterior angle is equal to the interior and opposite angle).
Therefore triangle EAB is equal to the triangle FDC by Proposition 4 (side-angle-side).

Subtract the triangle DGE from both triangles EAB and FDC.
The resulting trapezoids ABGD and EGCF are equal.
Add triangle GBC to each trapezoid.

Therefore the resulting parallelograms ABCD and EBCF are equal.

Q.E.D.

In the next proposition we will extend Proposition 35 to include bases that are not the same but are equal – George

Book 1 – Proposition 34

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

Let ACDB be a parallelogram, and BC be its diameter.

We will first prove that AB is equal to CD, AC is equal to BD, angle BAC is equal to CDB, and angle ABD is equal to ACD.

Since AC is parallel to BD and the line BC falls upon them, the alternate angles ABC and BCD are equal to one another by Proposition 29. (If a straight line falls upon two parallel lines, then the alternate angles are equal to one another.)
By the same reasoning, the alternate angles ACB and CBD are equal to one another.
So triangles ABC and DCB have angle ABC equal to angle BCD, angle ACB equal to angle CBD, and share the common side BC between those angles. By Proposition 26 (angle-side-angle) the remaining sides are equal to the remaining sides and the remaining angle is equal to the remaining angle.
Side AB is equal to side CD, side AC is equal to side CD, and angle BAC is equal to angle CDB.
All that remains is to prove that angle ABD is equal to angle ACD.
Angle ABC is equal to angle BCD and angle CBD is equal to is equal to angle ACB.
Thus angle ABD (ABC & CBD) is equal to angle ACD (BCD & ACB).

Therefore the opposite sides and angles are equal to one another.

Now we will prove that the diameter bisects the areas.

We will look to triangles ABC and BCD.
Side AB is equal to side CD, and side BC is common to the two triangles. The angles contained between the equal sides, ABC and BCD, are equal.
By Proposition 4 (side-angle-side), the triangle ABC is equal to the triangle DCB.

Therefore the diameter BC bisects the parallelogram ACDB.

Q.E.D.

The next proposition also involves parallelograms, comparing two parallelograms on the same base and in the same parallels.

Book 1 – Proposition 33

The straight lines joining equal and parallel straight lines [at the extremities which are] in the same directions [respectively] are themselves also equal and parallel.
[Desmos graphs can be found here]

Let AB and CD be equal and parallel, and let the straight lines AC and BD join them at the extremities in the same directions as shown.

We will prove that AC and BD are equal and parallel.
Let BC be joined, creating triangles ABC and BCD.

BC is a straight line that falls on the parallel lines AB and CD, so the alternate angles ABC and BCD are equal by Proposition 29 (a straight line that falls on two parallel lines makes alternate angles that are equal). Side AB is equal to side CD by hypothesis, and side BC is common to the triangles ABC and BCD.

So, by Proposition 4 (side-angle-side), the base AC is equal to the base BD and triangle ABC is equal to the triangle DCB. Therefore the angle ACB is equal to the angle CBD.

Since BC falls on the two straight lines AC and BD and the alternate angles ACB and CBD are equal, then by Proposition 27 (if a straight line that falls on two straight lines makes the alternate angles equal then the two lines are parallel to each other) we have that AC is parallel to BD.

So, we have proved that AC is equal to BD, and AC is parallel to BD.

Q.E.D.

ABCD is a parallelogram: opposite sides are equal and parallel. In the next proposition we will prove that the opposite sides and opposite angles in a parallelogram are equal – George

Book 1 – Proposition 32

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.
[Desmos graphs can be found here]

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

We will prove that (i) the exterior angle ACD is equal to the two interior and opposite angles ABC and CAB, and (ii) the three interior angles ABC, BCA, and CAB are equal to two right angles.

Using Proposition 31 (which covered how to construct a line through a given point that is parallel to a given line), let CE be drawn through point C that is parallel to AB.

Line AC falls upon the two parallel lines AB and CE, so by Proposition 29 (a straight line falling on two parallel lines makes the alternate angles equal to one another) the alternate angles BAC and ACE are equal to one another.

Line BD also falls upon the two parallel lines AB and CE, so Proposition 29 (exterior angle is equal to interior and opposite angle) tells us that the exterior angle ECD is equal to the opposite and interior angle ABC.

Angle ACD is made up of the angles ACE (which is equal to angle BAC) and ECD (which is equal to angle ABC). So, the exterior angle ACD is equal to the two interior and opposite angles ABC and BAC.

Moving on to the proof of (ii), add the angle ACB to both angle ACD as well as to the angles ABC and BAC.
Angles ACD and ACB are equal to the angles ABC, BAC, and ACB.
By Proposition 13 (a straight line set up on a straight line makes two angles that are equal to two right angles) angles ACD and BAC are equal to two right angles.

Therefore the three interior angles ABC, BAC, and ACB are equal to two right angles.

Q.E.D.

That’s two-thirds of the way through Book 1. Comments or questions? I’d love to hear them – George

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.