construction – uCLID

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 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 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 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 23

On a given straight line and a point on it to construct a rectilineal angle equal to a given rectilineal angle.
[Desmos graphs can be found here]

We begin with a line AB (let A be the point on it) and an angle DCE where D and E are selected at random.

The goal is to construct an angle on AB at point A that is equal to angle DCE.

Begin by joining DE. Construct a triangle AFG on AB such that AF, FG, and GA are equal to CD, DE, and EC respectively. (Proposition 22 shows how such a triangle can be constructed with sides equal to those 3 line segments.)

The two sides DC and CE are equal to the two sides FA and AG, and the base DE is equal to the base FG. By Proposition 8 (Two triangles with two sides equal to two sides respectively and equal bases have equal angles between the two sides) angles DCE and FAG are equal.

Thus we have constructed an angle on AB that is equal to the given angle.

Book 1 – Proposition 12

Proposition 12

To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.
[Link to graphs on Desmos]

Before we begin, here is the definition for right angles and perpendicular lines:
When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.To show that two lines are perpendicular we must show that the two lines form right angles.

Here is an infinite straight line AB and a point C that is not on it. The goal is to draw a line from point C that is perpendicular to AB.

Let point D be any point on the other side of AB from C, and draw a circle whose radius is equal to CD. I have labeled the circle as EFG, where points E and G are the points where the circle intersects the line AB.

Let the point H be the point at which the segment EG is bisected. (Proposition 10 covered how to find the point that bisected a line segment.)
Draw the segments from point C to the points G, H, and E.

We will now focus on triangles GCH and ECH as we try to show that the segment CH is perpendicular to AB.

GH is equal to HE, since H bisected EG.
CH is common to both triangles.
CG is equal to CE, because they are both on the circle with center C and radius CD.

So, by Proposition 8 (if two triangles have two equal sides and equal bases, then the angles between the equal sides are equal to each other, or in today’s introductory geometry classes: side-side-side implies equal angles), the two angles GHC and EHC are equal to each other.

Since the line segment CH set up on the straight line AB makes two angles GHC and EHC that are adjacent angles and equal to each other, both angles GHC and EHC are right angles.

By definition, the line segment CH is perpendicular to the line AB.

If you like what you see, or if you have suggestions for improvement, I would love to hear from you. Please leave any feedback as a comment – George

Book 1 – Proposition 11

Proposition 11

To draw a straight line at right angles to a given straight line from a given point on it.

[Desmos graphs are located here]

We will start with a line segment AB, and let C be the given point on it.

We will draw a line from point C at right angles from the segment AB.
We begin by plotting a point D on AC.
Let E be the point on CB such that CE is equal to CD. (Proposition 3 states that we can cut off a line segment that is equal to CD.)

On DE we can construct the equilateral triangle FDE (Proposition 1 shows the construction that shows an equilateral triangle can be built on any line segment), and join points F and C as shown.

We will now show that FC forms right angles with AB at point C by looking at triangles DCF and ECF.

DC is equal to EC by the way we chose point E.
CF is common to both triangles.
DF is equal to FE because they are both sides of the equilateral triangle DFE.

Therefore by Proposition 8 (if two triangles have three equal sides then the angles between two equal sides are equal), the angles DCF and ECF are equal to each other.

One of Euclid’s definitions will help us here: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the angles is right.

The two adjacent angles DCF and ECF are formed when a straight line (CF) is set up on a straight line (AB) and are equal to each other, so each of the angles is right.

This proposition will be used in the next proposition, a construction of drawing a line through a point that is perpendicular to a given straight line.

Questions, comments, or feedback? Please leave a comment below – George

Book 1 – Proposition 10

Proposition 10

To bisect a given straight line.

[Desmos.com graphs for this proposition]

Suppose we start with a line segment AB.

Construct the equilateral triangle ABC on it. (Proposition 1 covers how to construct an equilateral triangle om a line segment.)

Let angle ACB be bisected by the line segment CD, where the point D lies on the line segment AB. (Proposition 9 covers how to bisect a rectilinear angle.)

We will now compare the two triangles ACD and BCD.

AC is equal to BC because they are two sides of the equilateral triangle ABC.
Side CD is common to both triangles.
Angle ACD is equal to angle BCD, since CD bisects the angle ACB.

Since the two triangles have two equal sides, with equal angles between those equal sides, then the two bases AD and BD must be equal. (Proposition 4 states that if two triangles have equal side-angle-side, then the third sides must also be equal.)

Therefore AB has been bisected at D.

If you have any questions or feedback I’d love to hear it. Please leave a comment – George

Book 1 – Proposition 9

Proposition 9

To bisect a given rectilinear angle.

[Desmos graphs for this proposition]

This proposition is a construction that shows how to bisect an angle. We begin with the angle BAC.

Choose a point D at random on AB and cut off a segment AE from AC that is equal to AD. (Proposition 3)
Join DE.

Construct the equilateral triangle DEF be constructed using a point F on the opposite side of DE from point A . (Proposition 1)

Let AF be joined.

Now we will show that AF bisects the angle BAC by using the triangles DAF and EAF.

AD is equal to AE, since AE was cut off AC to equal AD.
AF is common to both triangles.
DF is equal to EF, since they are two of the sides of an equilateral triangle.

The conditions for Proposition 8 have been met, so angles DAF and EAF are equal to each other. Therefore the angle BAC has been bisected.

Now that we have learned how to bisect a rectilinear angle, we will learn how to bisect a given straight line (segment) in the next proposition.
As always, questions and feedback is welcome – George