January 2016 – uCLID

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 25

If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

We will start with two triangles ABC and DEF where AB is equal to DE and AC is equal to DF, but the base BC is greater than the base EF.

We will prove that angle BAC is greater than angle EDF. Suppose that angle BAC was not greater than angle EDF, then either it is equal to it or less than it.

Suppose that angle BAC is equal to angle EDF.

Then, by Proposition 4 (if two triangles have two equal sides and contain an equal angle between them, then the third sides must be equal), the base BC would have to be equal to the base EF, which it is not.

Suppose that angle BAC is less than angle EDF.

This is impossible by Proposition 24, which would imply that BC is less than EF, which it is not.
(Proposition 24: If two triangles have two equal sides but have one of the contained angles greater than the other, then the base of that triangle is greater than the other base.)

So, since angle BAC is not equal to or less than angle EDF, it must be greater than angle EDF.

Q.E.D.

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

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 22

Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. (Proposition 20)
[Desmos step-by-step graphs can be found here]

Let three lines A, B, and C satisfy the condition that any two taken together are greater than the remaining one.

We will now construct a triangle whose sides are equal to these three lines.
Set out a straight line beginning at a point D that continues through point E.
Place a point F such that DF is equal to A, a second point G such that FG is equal to B, and a third point H such that GH is equal to C.
(Proposition 3 allows us to create a line segment of length equal to the length of a given line segment.)

Construct a circle centered at F with a radius equal to DF. This circle can be labeled as DKL.

Construct a second circle centered at G with a radius equal to GH. This circle can be labeled as HKL, where K and L are the two points of intersection between the two circles.

Let FK and GK be joined.

We will now show that triangle FGK has three sides that are equal to A, B, and C.

Since K is on the circle centered at F, FK is equal to the radius DF which is also equal to A. Therefore FK is equal to A.
Since K is also on the circle centered at G, GK is equal to the radius GH which is also equal to C. Therefore GK is equal to C.
FG was set up to be equal to B.

FK, FG, and GK are equal to the three straight lines A, B, and C.

Therefore a triangle with sides equal to the three lines has been constructed.

Does triangle FGL also satisfy the requirements? As an exercise, see if you can show that triangle FGL has sides that are the lengths of A, B, and C.

The next proposition is another construction, requiring us to construct an angle that is equal to a given angle. The construction in Proposition 22 plays a large role in that construction.

George

Book 1 – Proposition 21

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
[Desmos graphs can be found here]

Just a quick translation here – If you start with a triangle, and from its base construct a new triangle whose third vertex is located within the original triangle, then the two new sides will be less than the original two sides but the angle contained between them will be greater.

We will start with a triangle ABC, and construct a triangle BCD on the base BC such that D is within the triangle.

We will prove that the two sides BD and DC are less than BA and AC, and that the angle BDC is greater than the angle BAC.

Let BD be drawn through to a point E that lies on AC.

In triangle ABE, Proposition 20 (two sides of a triangle together are greater than the remaining side) tells us that sides AB and AE together are greater than BE.
If we add EC to each, we have that AB and AC (AE & EC) together are greater than BE and EC.

We now have that AB and AC greater than two other segments (BE and EC). We need to show that those two sides are also greater than BD and DC.

Looking to triangle CDE, Proposition 20 tells us that the two sides CE and ED together are greater than DC.
If we add BD to each we have that EC and BE (BD & ED) together are greater than DC and BD.

Therefore AB and AC together are greater than BD and DC.
[ AB & AC > BE & EC > BD & DC ]

Now we move on to prove the statement concerning the angles.
(Angle BDC is less than angle BAC)

We will begin by using Proposition 16 (exterior angle of a triangle is greater than either of the opposite interior angles) in triangle CDE.

Angle BDC (exterior angle) is greater than angle CED (an opposite interior angle). Angle CED is equal to angle CEB, so we have angle BDC as greater than angle CEB.

In triangle ABE, the exterior angle CEB is greater than the opposite interior angle BAE (which is equal to angle BAC).

Since angle BDC is greater than angle CEB, which is greater than angle BAC, therefore angle BDC is greater than angle BAC.

Q.E.D.

Book 1 – Proposition 20

In any triangle two sides taken together in any manner are greater than the remaining one.
[Desmos graphs available here]

Intuitively this should make sense, because if the two sides were shorter than the third they would not even cover the third segment laid end to end, making it impossible to form a triangle.

For a triangle ABC, we will prove the sides AB and AC are greater than side BC. A similar strategy can be used to show that AB and BC are greater than AC, and that AC and BC are greater than AB.

Extend the segment BA through a point D such that the segment AD has the same length as segment AC. (Proposition 2 allows us to construct a line segment of the same length as AC.)
Let DC be joined.

Look at triangle ACD. Since AD is equal to AC, triangle ACD is an isosceles triangle. By Proposition 5, angle ADC is equal to angle ACD. (The base angles of an isosceles triangle are equal.)

Because angle BCD is greater than angle ACD, we know that angle BCD is greater than angle ADC. We also know that angles ADC and BDC are equal.

Looking at triangle BCD …
Since angle BCD is greater than angle BDC, side BD is greater than side BC by Proposition 19 (the greater angle is subtended by the greater side).
The side BD is equal to AD and AB.
Since AD was created equal to AC, side BD is equal to AC and AB.
Thus AC and AB together are greater than BC.
[ BD > BC … AD & AB together > BC … AC & AB > BC]

Q.E.D.

We showed that two arbitrarily selected sides together are greater than the remaining side, so our results apply to each pairing in the triangle.

Questions? Comments? You know what to do – George

Book 1 – Proposition 19

In any triangle the greater angle is subtended by the greater side.

We will start with a triangle ABC in which angle ABC is greater than angle ACB.

We will now prove that side AC is greater than side AB.

Suppose that this was not true. In other words, AC is not greater than side AB. Then AC would either be equal to AB or less than AB.

AC equal to AB?
According to Proposition 5 (the two angles at the base of an equilateral triangle are equal) AC cannot be equal to AB, because they would be two equal sides in an isosceles triangle and that would imply that angles ABC and ACB would be equal which they are not.

AC less than AB?
AC cannot be less than AB, for that would violate Proposition 18 (in any triangle the greater side subtends the greater angle) in that the greater side (AB) does not subtend the greater angle (ABC).

Therefore AC is greater than AB.

Q.E.D.

Questions? Comments? Please leave feedback – George

Book 1 – Proposition 18

In any triangle the greater side subtends the greater angle.
[Desmos graphs can be found here]

Let ABC be a triangle with aide AC greater than AB.

We will prove that angle ABC is greater than angle ACB.

Since AC is greater than AB, put a point D on AC such that AD is equal to AB.

Angle ADB is an exterior angle of triangle BCD, so by Proposition 16 it is greater than the opposite interior angle DCB.
(Proposition 16 stated that the exterior angle is greater than either of the opposite interior angles.)

Angle ADB is equal to angle ABD, because AB and AD are two equal sides in a triangle.
So, angle ABD is greater than angle ACB (which is the same as angle DCB).
Since angle ABC is greater than angle ABD, angle ABC is also greater than angle ACB.
[ ABC > ABD = ADB > DCB = ACB ]

Q.E.D.

We just proved that the greater side subtends the greater angle.
In Proposition 19 we will prove that the greater angle is subtended by the greater side.

I appreciate any feedback or questions – George