February 2016

Book 1 – Proposition 45

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

Starting with a given rectilineal figure ABCD and a given angle E, we will construct a parallelogram equal to ABCD in an angle equal to E.

Join BD to express ABCD as two triangles: ABD and DBC.
Use Proposition 42 (construct a parallelogram in a given angle that is equal to a given triangle) to construct a parallelogram FGHK equal to triangle ABD in an angle FKH that is equal to angle E.

Use Proposition 44 (to construct a parallelogram to a given line in a given angle that is equal to a given triangle) to construct a parallelogram GHML on the line GH equal to triangle DBC in the angle GHM that is equal to angle E.

Since they are both equal to angle E, angles FKH and GHM are equal to one another.
Add angle KHG to both, so angles FKH and KHG together are equal to the angles GHM and KHG.
By Proposition 29 (a straight line falling on two parallel lines creates two interior angles on the same side that are together equal to two right angles) angles FKH and KHG together are equal to two right angles, so angles GHM and KHG are also together equal to two right angles.
So the straight line GH at a point H meets two straight lines KH and HM (that are not lying on the same side of GH) and makes two alternate angles equal to two right angles. By Proposition 14 (if with a straight line, and at a point on it, two straight lines not on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with each other), KH is in a straight line with HM.

The straight line GH falls upon two parallel lines KM and FG. By Proposition 29 (a straight line falling on two straight lines creates interior angles on the same side that together are equal to two right angles) the alternate angles HGF and GHM are equal to one another.
Add angle HGL to both, so the angles HGF and HGL together are equal to the angles GHM and HGL.
Again, by Proposition 29, the angles GHM and HGL together are equal to two right angles. And so the angles HGF and HGL are together equal to two right angles.
So, by Proposition 14, FG is in a straight line with GL (since GH meets the two lines that are not on the same side and makes two alternate angles equal to two right angles).

Since by Proposition 34 (opposite sides of a parallelogram are equal) FK is equal to GH, and GH is equal to LM, therefore FK is equal and parallel to LM by Proposition 30 (straight lines parallel to the same straight line are parallel to one another).
The lines FL and KM join the lines FK and LM at their extremities, so by Proposition 33 (the straight lines joining equal and parallel straight lines at the extremities which are in the same direction are themselves also equal and parallel) FL and KM are also equal and parallel.

Therefore FLMK is a parallelogram, and it is equal to the two parallelograms FGHK and GHML together.

Since FGHK is equal to triangle ABD and GHML is equal to the triangle DBC, the parallelogram FLMK is equal to the two triangles ABD and DBC and therefore is equal to the rectilineal figure ABCD.

Since FLMK is in the angle FKM and that angle is equal to angle E, we have constructed a parallelogram equal to the given figure in the given angle.

Whew – George

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 40

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

Let ABC and CDE be equal triangles with equal bases BC and CE.

We will prove that AD is parallel to BE to show that the two triangles are in the same parallels.
Begin by joining AD.

If AD is not parallel to BE, then draw AF through point A so that AF is parallel to BE. (Proposition 31 covers drawing a line through a given point that is parallel to a given line.) Join FE as well.

Triangles ABC and FCE have equal bases (BC and CE) and in the same parallels (BE and AF), so by Proposition 38 (triangles on equal bases and in the same parallels are equal to one another) triangles ABC and FCE are equal to one another.
Since triangle ABC is also equal to triangle CDE, triangle FCE must also be equal to triangle CDE which is impossible. (One of those triangles must be greater than the other.)

Therefore AF is not parallel to BE.

Therefore AD is parallel to BE, and the two triangles are in the same parallels.

Q.E.D.

The next proposition will compare triangles and parallelograms with equal bases in the same parallels – George

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