parallelogram

Book 1 – Proposition 46

On a given straight line to describe a square.
[Desmos graphs can be found here]

Let AB be the given straight line. We will construct a square on AB.

Use Proposition 11 (constructing a straight line at a right angle to a given line at a point) to draw AC at a right angle to AB at the point A.

Select point D on AC such that AD is equal to AB.
Using Proposition 31 (covers drawing a straight line through a point that is parallel to a given line), through point D draw DE that is parallel to AB.
Also, through point B draw BE that is parallel to AD.

ADEB is a parallelogram, and by Proposition 34 (in a parallelogram opposite sides are equal and opposite angles are equal) AB is equal to DE and AD is equal to BE.

Since AB is equal to AD, all four sides of the parallelogram are equal to each other. Therefore the parallelogram ADEB is equilateral.

Now we will show that the parallelogram is also right-angled, making it a square.
Since AD falls upon the parallel lines AB and DE, Proposition 29 (if a line falls on two parallel lines, then the interior angles on the same side are equal to two right angles) tells us that the two angles BAD and ADE are together equal to two right angles.
But angle BAD is a right angle, so angle ADE is also a right angle.
Using Proposition 34 again, the two angles opposite BAD and ADE are equal to those two angles. So, angles ABE and BED are also each right angles.

Since ADEB is right-angles and equilateral, it is a square described on the straight line AB.

Next proposition … the Pythagorean Theorem!

Book 1 – Proposition 44

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]

Let AB be the given straight line, C be the given triangle, and D the given angle.

We will apply a parallelogram to AB in an angle equal to D that is equal to the triangle C.
Use Proposition 42 (covers how to construct a parallelogram, in a given angle, that is equal to a given triangle) to construct the parallelogram BEFG that is equal to the triangle C in the angle EBG that is equal to D. Place the parallelogram so that BE is in a straight line with AB.

Let FG be drawn through to a point H such that AH is parallel to BG and EF. (Use Proposition 31: through a given point to draw a straight line that is parallel to a given line.)

Join HB.

Since the straight line HF falls on the parallel lines AH and EF, Proposition 29 (a straight line falling on two parallel lines creates two interior angles on the same side that are equal to two right angles) tells us that the angles AHF and HFE together are equal to two right angles.
Therefore the two angles BHG and GFE are less than two right angles.
(BHG is contained in and therefore less than AHF, and angle GFE is equal to HFE.)
Straight lines produced indefinitely from angles less than two right angles must meet, so the line through HB must meet the line through FE.
Let HB and FE be produced and meet at a point K.
Through K use Proposition 31 to draw a line KL that is parallel to FH and EA.
Produce HA to the point L and produce GB to a point M that lies on LK.

HLKF is a parallelogram, and HK is its diameter.
ABGH and MKEB are parallelograms contained in HLKF.
LMBA and BEFG are the complements about HK, and by Proposition 43 (in any parallelogram the complements of the parallelograms about the diameter are equal to one another) the two parallelograms LMBA and BEFG are equal.

Parallelogram BEFG was constructed to be equal to triangle C, so parallelogram LMBA is also equal to the triangle C.

Angle ABM is equal to angle GBE by Proposition 15 (if two straight lines intersect each other, they make the vertical angles equal to one another), and since angle GBE was constructed equal to angle D then the angle ABM is equal to angle of D.

Therefore parallelogram LMBA equal to the given triangle C has been applied to the given straight line AB, in the angle ABM equal to the given angle D.

The next proposition involves constructing a parallelogram in a given angle that is equal to a given rectilineal figure (instead of a given triangle) – George

Book 1 – Proposition 43

In any parallelogram the complements of the parallelogram about the diameter are equal to one another.
[Desmos graphs can be found here]

Let ABCD be a parallelogram with diameter AC.

About AC let AEKH and KGCF be parallelograms.
Call BEKG and DHKF the complements.

We will prove that the parallelograms BEKG and DHKF are equal to one another.

Since ABCD is a parallelogram and AC its diameter, Proposition 34 (the diameter of a parallelogram bisects it) tells us that triangle ABC is equal to the triangle ACD.

The parallelogram AEKH has a diameter AK, so again by Proposition 34 triangle AEK is equal to the traingle AHK.

KC is a diameter of the parallelogram KGCF, so by Proposition 34 triangle KGC is equal to triangle KFC.

Triangle ABC is composed of triangle AEK, triangle KGC, and parallelogram BEKG.
Triangle ADC is composed of triangle AHK, triangle KFC, and parallelogram DHKF.
Triangles ABC and ADC are equal to each other, and removing the two sets of equal triangles from each (AEK = AHK and KGC = KFC), the complements (parallelograms BEKG and DHKF) must be equal to each other.

Therefore parallelogram BEKG is equal to parallelogram DHKF. The complements of the parallelograms about the diameter are equal to one another.

Q.E.D.

The next proposition is a construction that relies on the result in Proposition 43 – George

Book 1 – Proposition 42

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]

Start with a given triangle ABC and a given rectilineal angle D.

The goal is to construct a parallelogram that is equal to the triangle ABC in an angle equal to D.

Let E bisect BC. Join AE. Use Proposition 23 (Construction of an angle on a straight line that is equal to a given angle) to construct an angle CEF that is equal to angle D.

Draw AG through A that is parallel to EC. Draw CG through C that is parallel to EF. These lines can be drawn using Proposition 31 (Construction of a straight line through a given point that is parallel to a given line).

Since FG is parallel to EC and EF is parallel to CG, FECG is a parallelogram.

Now we look at triangles ABE and AEC.
Since BE is equal to EC (as E bisects BC), the two triangles have equal bases.
The two triangles are in the same parallels (BC and AG).
By Proposition 38 (two triangles on equal bases and in the same parallels are equal to one another) triangle ABE is equal to triangle AEC.

Since the triangle ABC is composed of the two equal triangles ABE and AEC, the triangle ABC is double the triangle AEC.

But, by Proposition 41 (if a parallelogram has the same base as a triangle and they are in the same parallels then the parallelogram is double the triangle) the parallelogram FECG is double the triangle AEC since they share the same base (EC) and they are in the same parallels.

Since both the parallelogram FECG and the triangle ABC are double the triangle AEC, the parallelogram FECG is equal to the triangle ABC.
The angle FEC is equal to the angle D.
We have constructed the parallelogram that meets the necessary conditions.

Book 1 – Proposition 41

If a parallelogram has the same base as a triangle and is in the same parallels, then the parallelogram is double the triangle.
[Desmos graphs can be found here]

We begin with parallelogram ABCD and triangle EBC sharing the same base BC and be in the same parallels BC and AE.

We will prove that the parallelogram ABCD is double the triangle EBC.
Start by joining AC.

Compare triangles ABC and EBC. They have the same base BC, and they are in the same parallels BC and AE, so the two triangles are equal by Proposition 37. (Triangles with the same base that are in the same parallels are equal to one another.)

Since AC is the diameter of the parallelogram ABCD, the parallelogram is double the triangle ABC by Proposition 34. (The diameter of a parallelogram bisects it.)

Since the parallelogram ABCD is double the triangle ABC, and the triangle ABC is equal to the triangle EBC, therefore the parallelogram ABCD is double the triangle EBC.

Q.E.D.

In the next proposition we return to a construction of a parallelogram that is equal to a given triangle – 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.