desmos – Page 2

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 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

Book 1 – Proposition 27

If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.
[Desmos graphs can be found here]

Let line EF fall on the straight lines AB and CD with the alternate angles AEF and EFD equal to one another.

We will prove that AB is parallel to CD by showing that they cannot intersect. Suppose that they did intersect at a point G in the direction of B and D.

We will look at triangle GEF. By Proposition 16 (In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles) the exterior angle AEF must be greater than the interior angle EFG. But, by hypothesis, those two angles are equal which produces a contradiction.

Therefore AB and CD do not meet at a point in the direction of B and D. In a similar fashion it can be shown that they do not intersect at a point in the direction of C and D.

Thus, AB is parallel to CD.

Q.E.D.

This proposition will be used in the proof of Proposition 28 which states that a line that intersects two parallel lines produces alternate angles that are equal to one another.

Book 1 – Proposition 26

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjourning the equal angles, or that subtending one of the equal angles, they will have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.
[Desmos graphs can be found here]

So if two triangles have two pairs of equal angles and an equal side, we will prove that the remaining sides and angles are equal as well. This can be done by showing that the two triangles are equal.

Start with the two triangles ABC and DEF. Angle ABC is equal to angle DEF and angle BCA is equal to angle EFD.

Part 1: Let the sides adjoining the equal angles be equal, in other words, BC is equal to EF.
We will prove that AB is equal to DE, AC is equal to DF, and angle BAC is equal to angle EDF.

If AB is not equal to DE, let AB be greater than DE since one of them must be greater. Let G be a point on AB such that BG is equal to DE, and join CG.

We will compare triangle BCG to triangle DEF.

BG is equal to DE (by the way we chose G)
BC is equal to EF (those two sides were given to be equal)
Angle GBC is equal to angle DEF (since angle GBC is equal to angle ABC)

Therefore, by Proposition 4 (side-angle-side), the base GC is equal to the base DF, the triangle BCG is equal to the triangle DEF, and angle GCB is equal to angle DFE.
But angle DFE is equal to angle ACB by hypothesis, which implies that angle GCB is equal to angle ACB. That is not possible. Thus side AB must be equal to side DE.

Returning to comparing triangles ABC and DEF …
We have the two sides AB and BC equal to the two sides DE and EF, and the angles contained between them (ABC and DEF) are equal.
Again using Proposition 4, the two bases AC and DF are equal and the remaining angles BAC and EDF are equal as well.

Part 2: Let two sides subtending equal angles be equal. We will choose that AB is equal to DE. (We could also choose that AC is equal to DF.)
We will prove that BC is equal to EF, AC is equal to DF, and angle BAC is equal to angle EDF.

If BC is not equal to EF, let BC be greater than EF since one of them must be greater. Let H be a point on BC such that BH is equal to EF, and join AH.

We will compare triangle ABH to triangle DEF.

BH is equal to EF (by the way we chose H)
AB is equal to DE (those two sides were given to be equal)
Angle ABH is equal to angle DEF (since angle ABH is equal to angle ABC)

Once again, by Proposition 4 (side-angle-side), the base AH is equal to the base DF, the triangle ABH is equal to the triangle DEF, and angle AHB is equal to angle DFE.
But angle DFE is equal to angle ACB by hypothesis, which implies that angle AHB is equal to angle ACB. That is not possible. Thus side BC must be equal to side EF.

Returning to comparing triangles ABC and DEF …
We have the two sides AB and BC equal to the two sides DE and EF, and the angles contained between them (ABC and DEF) are equal.
By Proposition 4, the two bases AC and DF are equal, triangles ABC and DEF are equal, and the remaining angles BAC and EDF are equal as well.

Q.E.D.

Notice that the proofs in parts 1 and 2 are very similar.

Book 1 – Proposition 24

If two triangles have the two sides equal to the two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.
[Desmos graphs can be found here]

We start with two triangles ABC and DEF, for which side AB is equal to side DE, side AC is equal to side DF, and the angle at A is greater than the angle at D.

We will prove that side BC is greater than side EF.

Since angle BAC is greater than angle BAC, by Proposition 23 we can construct an angle EDG on the straight line DE that is equal to angle BAC.
(Proposition 23 covers the construction of an angle equal to a given angle.)
Choose the point G so that DG is equal to both AC and to DF. Join EG and FG.

Since AB is equal to DE and AC is equal to DG and angle BAC is equal to side EDG, then by Proposition 4 (side-angle-side) the base BC is equal to the base EG.

Since DF is equal to DG, angle DGF is equal to angle DFG by Proposition 5 (concerning equal angles in an isosceles triangle). Therefore angle DFG (being equal to angle DGF) is greater than angle EGF.

Since angle EFG is greater than angle DFG, then angle EFG is also greater than angle EGF.

Focus on triangle EFG.
Angle EFG is greater than angle EGF, so side EG is greater than side EF because Proposition 19 that the greater angle is subtended by the greater side.

But EG is equal to BC, so BC must also be greater than EF.

Q.E.D.

We are halfway through Book 1 – any comments are appreciated – George