euclid – Page 2

Book 1 – Proposition 39

Equal triangles which are on the same base and on the same side are also in the same parallels.
[Desmos graphs can be found here]

Let triangles ABC and DBC have the same base BC and both be on the same side of it.

Join AD. We will prove that AD is parallel to BC.

If AD is not parallel to BC, let A be joined to a point E on BD such that AE is parallel to BC. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join EC.

By Proposition 37 (triangles on the same base and in the same parallels are equal) triangle ABC is equal to triangle EBC as they share the same base (BC) and are in the same parallels.

Since triangle ABC is equal to triangle DBC, then triangle EBC is also equal to triangle DBC which is impossible.
Therefore AE cannot be parallel to BC.
The same is true for any other straight line except AD.
Therefore AD is parallel to BC.

Q.E.D.

In the next proposition we will extend “the same base” to “equal bases” – George

Book 1 – Proposition 38

Triangles which are on equal bases and in the same parallels are equal to one another.
[Desmos graphs can be found here]

Let triangles ABC and DEF be on equal bases BC and EF, and in the same parallels BF and AD.

We will prove that triangle ABC is equal to triangle DEF.
Let AD be produced in both direction to points G and H such that BG is parallel to CA and FH is parallel to ED. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.)

The parallelogram GBCA is equal to the parallelogram DEFH by Proposition 36 as they are on equal bases BC and EF and are in the same parallels BF and GH.
(Proposition 36 states that two parallelograms on equal bases that are in the same parallels are equal to each other.)

The diameter AB bisects the parallelogram GBCA, so the triangle ABC is half of the parallelogram GBCA by Proposition 34 (the diameter of a parallelogram bisects the areas).
By similar reasoning, Proposition 34 tells us that triangle DEF is half of the parallelogram DEFH since DF is a diameter of the parallelogram.

Triangles ABC and DEF are each half of two equal parallelograms, and therefore are equal to each other.

Q.E.D.

Book 1 – Proposition 37

Triangles which are on the same base and in the same parallels are equal to one another.
[Desmos graphs can be found here]

Let ABC and DBC be triangles on the same base BC and in the same parallels AD and BC.

We will prove that triangles ABC and DBC are equal.

Project the line AD in both directions to points E and F such that BE is parallel to CA and CF is parallel to BA.
Proposition 31 covered how to construct a line through a given point that is parallel to a given line.
(BE passes through B and is parallel to CA. CF passes through C and is parallel to BA.)

EBCA and DBCF are parallelograms. The two parallelograms share the same base (BC) and are in the same parallels (BC and EF), so by Proposition 35 they are equal.

By Proposition 34 (the diameter of a parallelogram bisects its area) triangle ABC is half of the parallelogram EBCA because its diameter AB bisects the parallelogram.

Again by Proposition 34, since DC is the diameter of the parallelogram DBCF, triangle DBC is half of the parallelogram DBCF.

Since the triangles ABC and DBC are each half of equal parallelograms EBCA and DBCF, the two triangles ABC and DBC are therefore equal to one another.

Q.E.D.

The next proposition extends Proposition 37 from “the same base” to “equal bases” – George

Book 1 – Proposition 36

Parallelograms which are on equal bases and in the same parallels are equal to one another.
[Desmos graphs can be found here]

Consider the parallelograms ABCD and EFGH which are on equal bases BC and FG and in the same parallels AH and BG as shown.

We will prove that the parallelogram ABCD is equal to the parallelogram EFGH.
Let BE and CH be joined.

Since BC is equal to FG (hypothesized equal bases) and FG is equal to EH by Proposition 34 (in a parallelogram, opposite sides and opposite angles are equal), then BC is equal to EH.
BC is also parallel to EH, and they are joined by BE and CH.
By Proposition 33 (straight lines that join equal and parallel lines are themselves also parallel and equal), BE and CH are equal and parallel.
Therefore EBCH is a parallelogram by Proposition 34.

The parallelogram EBCH is equal to the parallelogram ABCD by Proposition 35 (parallelograms on the same base and in the same parallels are equal to one another) because they have the same base BC and are in the same parallels BC and AH.

Also, by Proposition 35, the parallelogram EBCH is equal to the parallelogram EFGH as they share the base EH and are in the same parallels EH and BG.

Since both ABCD and EFGH are equal to EBCH, they are equal to each other. Therefore the parallelogram ABCD is equal to the parallelogram EFGH.

Q.E.D.

In the next proposition we will prove a similar result for triangles – George

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