Taking A Short Break

Now that I have made it to the end of Book 1, I am going to take a short break before moving on to Book 2. I will use that time to add comments and insights to the 48 Propositions in Book 1.

If you have any comments on things that could be added to those first 48 please leave me a comment.

You can follow me @uCLID_blog on Twitter to keep up with my progress and be notified when I move on to Book 2.

Thanks – George

Book 1 – Proposition 48

If in a triangle the square one one of the sides is equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.
[Desmos graphs can be found here]

For the triangle ABC let the square on side BC be equal to the squares on side AB and BC.

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We will prove that angle BAC is a right angle.
Let AD be drawn from point A at a right angle to AC.
Let AD be made equal to AB.
Join DC.

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Since DA is equal to AB, the square on DA is equal to the square on AB.
Add the square on AC to each.
The sum of the squares on DA and AC is equal to the sum of the squares on AB and AC.
The square on DC is also equal to the sum of the squares on DA and AC because the angle DAC is a right angle. (Proposition 47: In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.)
Since the square on BC is equal to the squares on AB and AC, the square on BC is also equal to the square on DC. Therefore the side DC is equal to the side BC.

Now compare triangle DAC to triangle BAC.
Side DA was constructed to be equal to side AB, side AC is common, and the two bases DC and BC were shown to be equal.
Therefore by Proposition 8 (side-side-side) triangle DAC is equal to triangle BAC and angle DAC is equal to angle BAC.
Since angle DAC is a right angle, therefore angle BAC is a right angle.

Q.E.D.


Proposition 47 showed that for a right triangle that the square on the side subtending the right angle is equal to the squares on the sides containing that right angle. Proposition 48 takes us in a different direction. If the square on one side is equal to the squares of the other two sides, then the angle opposite that one side must be a right angle. Notice how the two propositions work in the opposite direction of each other.

We have reached the end of Book 1 of Euclid’s Elements. After a short break I will continue on with Book 2. Thanks for reading – George

Book 1 – Proposition 47

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
[Desmos graphs can be found here]

Let ABC be a right-angled triangle with angle BAC being a right angle.

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We will prove that the square on BC is equal to the squares on AB and AC.
Use Proposition 46 (the construction of a square on a straight line) to construct the square BDEC on the line BC, the square AGFB on the line AB, and the square AHKC on the line AC.

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Through the point A draw the line AL that is parallel to BD and CE.
Join AD and CF.

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Angles BAC and BAG are each right angles. The straight line BA falls on the two straight lines AC and AG on different sides of it. The adjacent angles are equal to two right angles, so GA is in a straight line with AC by Proposition 14 (if any line and any two lines not on the same side of it form two adjacent angles that are equal to two right angles, then the two lines will be in a straight line with one another).

Also. by the same reasoning, BA is in a straight line with AH.

The right angle DBC is equal to the right angle FBA.
Add angle ABC to each. Therefore angle DBA is equal to angle FBC.
SIde DB is equal to side BC (on the same square), and side AB is equal to side FB.
Comparing triangles ABD and FBC: Sides AB and BD are equal to sides FB and BC and the angles between the two sides are also equal. So, by Proposition 4 (side-angle-side) the base AD is equal to the base FC and the triangle ABD is equal to the triangle FBC.

By Proposition 41 (if a parallelogram and a triangle share the same base and are in the same parallels, then the parallelogram is double the triangle) the parallelogram BL is double the triangle ABD.

Also by Proposition 41, the square AGFB is double the triangle FBC as they share the same base (FB) and are in the same parallels (FB and GC).

Since the parallelogram BL is double the triangle ABD and the square AGFB is double the triangle FBC, and those two triangles are equal to each other, the parallelogram BL is equal to the square AGFB.

Join AE and BK.

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In a similar fashion we can prove that the parallelogram CL is equal to the square AHKC.

The whole square BDEC is equal to the parallelograms BL and CL, which are equal to the two squares AGFB and AHKC.

The square BDEC is described on BC, the square AGFB is described on AB, and the square AHKC is described on AC.

Therefore the square on BC (subtending the right angle) is equal to the squares on AB and AC (the sides containing the right angle).

Q.E.D.


This Proposition is equivalent to the Pythagorean theorem. If the lengths of the two legs of a triangle are a and b and the hypotenuse is c, then a^2 + b^2 = c^2. In this Proposition and proof you can actually see the squares instead of working with an abstract formula. There are many proofs of the Pythagorean theorem – I encourage you to look up some of the others.
George

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.

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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.

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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.

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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 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.

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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.

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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.

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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.

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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).

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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.

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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.

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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.)

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Join HB.

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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.

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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.

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

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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.

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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.

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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).

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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.

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

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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.

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

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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.

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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