1. Euclid’s Proposition 32 and Proclus’s Commentaries
A Geometry lesson in English
Task sheet 1
A. Watch the following video about ancient Greek mathematicians. Write down
the names that you hear.
______________________________________________________________
B. The following quote comes from Albert Einstein. Can you guess the name of
the ancient Greek mathematician that is missing?
‘If ________________ did not kindle your youthful enthusiasm, you were
not born to be a scientific thinker’.
C. Geometry glossary. In pairs try to match the following words and phrases to
their meaning.
1. Extend a. άθροισμα
2. Angle b. ισούμαι
3. Degree c. προεκτείνω
4. Equal d. τέμνω
5. Right angle e. εσωτερικός
6. Sum f. εξίσωση
7. Equation g. ορθή γωνία
8. Interior h. εντός επί τα αυτά γωνίες
9. Proposition i. γωνία
10. Intersect j. πρόταση
11. Corresponding angles k. εντός εναλλάξ γωνίες
12. Alternate interior angles l.μοίρα
13. Interior angles m.εντός εκτός και επί τα αυτά γωνίες
D. Watch the following video about Euclid’s proposition 32 from Book 1.
Complete the gaps with the missing information.
This video presentation is going to be on proposition 32 of Book 1 which
states that if we have a triangle and we extend the base further out, that the
exterior angle will be equal to the sum of the two opposite interior angles
and that the sum of all the interior angles will equal two right angles. So, let’s
begin. First thing we’re going to do is to draw a parallel line to AB.
Now, according to proposition 29 we have two
1)_______________________AB and CE so the interior and opposite angles
will be equal. So ACE will equal CAB as equal to γ. Now if we look again at the
2. two parallel lines BA and CE, the interior and exterior angles are also equal.
So ABC is equal to ECD which is equal to α. Νοw if we look at our angle ACD,
it’s equal to δ, which is equal to γ plus α. If we look at our line BD, we know
that β plus δ is equal to two right angles according to 2) _______________.
So now we have δ is equal to γ 3)_____________ α and β plus δ is equal to
two right angles. So we have β plus γ plus α is equal to two right angles
having substituted this into that part of the equation. So here we have
demonstrated that for a triangle the 4) _____________ angle δ is equal to
the sum of the two interior angles and that the sum of all the interior angles
is equal to two right angles or 180 5) ________________.
Ερωτήσεις
1. Ποιο είναι το αντίστοιχο θεώρημα-πόρισμα στο σχολικό βιβλίο Ευκλείδειας
Γεωμετρίας;
2. Ποιες είναι οι ομοιότητες και διαφορές ανάμεσα στη διατύπωση και την απόδειξη
της πρότασης από τον Ευκλείδη με τις αντίστοιχες του σχολικού βιβλίου;
3. Ποιες είναι οι προτάσεις (θεωρήματα-πορίσματα) πάνω στις οποίες βασίζεται η
αποδεικτική διαδικασία του Ευκλείδη για την πρόταση 32;
3. Euclid’s Proposition 32 & Proclus’s Commentaries
A Geometry lesson in English
Task sheet 2
A. Read the comments made by Proclus on Euclid’s proposition. Working in
groups put the following subheadings in the correct place. Underline the
words/phrases that helped you.
One side of every angle being produced, the external angle of the triangle will
be equal to the internal and opposite angles ; and the three internal angles of
a triangle are equal to two right angles.
As much as was deficient in the sixteenth and fifteenth proposition, so much
Euclid adds in the present. For we not only learn by this theorem that the
external angle of a triangle is greater than either of the internal and opposite
angles, but likewise how much it is greater; since as it is equal to both it is
greater than any of the remaining angles. Nor do we alone know from this
theorem that any two angles of a triangle are less than two right but by how
much they are less; for they are deficient by the remaining third. The former,
therefore, were more indefinite theorems: but this brings with it, on both
sides, a boundary to science. We must not, however, call them on this
account superfluous: for they are of the greatest utility in many
demonstrations; and the present is proved by their assistance.
And besides this, it is necessary that our knowledge, proceeding from the
imperfect to the perfect, should pass from indeterminate apprehensions to
determinate and certain propositions.
But the institutor of the Elements, by drawing a parallel externally, exhibits
each of the objects of investigation. It is, however, possible that the same
thing may be shewn without drawing the parallel externally; and this, by only
changing the order of the things exhibited. For Euclid first shews that the
external angle is equal to the internal and opposite and from this he proves
the remainder. But we shall demonstrate this by a contrary mode of
proceeding.
a)Proclus’sproof b) Comparingpresentandformerpropositions
c)The Peripateticschool d)The proposition e) The
philosophyof the provingprocedure
4. Let there be then a triangle ABC,
and let the side BC be produced to the point E. Then take a point Z in BC,
and connect AZ, and through the point Z, let ZD be drawn parallel to AB.
Because, therefore, ZD is parallel to AB, and a right line AZ falls upon these
parallels, as also a right line BC, hence, the alternate angles are equal, and
the external is equal to the internal angle. The whole, therefore, AZC is equal
to ZAB, added to ABZ. In like manner we may shew by drawing a parallel,
that the angle AZB, is equal to the angles ZAC, ACZ. The two, therefore, AZB
and AZC, are equal to the three angles of the triangle: and, hence, the three
angles of a triangle are equal to two right, viz. to AZB, added to AZC. But ACZ ,
ACE, are also equal to two right angles. Let, therefore, the common angle
ACZ, be taken away: and then the remaining external angle will be equal to
the internal and opposite angles. And after this manner may the present
theorem be exhibited.
But Eudemus, the Peripatetic, ascribes the invention of this theorem to the
Pythagoreans, I mean that every triangle has its internal angles equal to two
right, and says that they demonstrate it in the following manner.
Let there be a triangle ABC, and
let there be drawn through the point A, a line DE, parallel to BC. Because,
therefore, the right lines DE, BC, are parallel, the alternate angles are equal.
Hence, the angle DAB, is equal to the angle ABC; and the angle EAC, to the
angle ACB. Let the common angle BAC, be added. The angles, therefore, DAB,
BAC, CAE, that is, the angles DAB, BAE, and that is two right, are equal to the
5. three angles of the triangle. And such is the demonstration of the
Pythagoreans.
B. According to Proclus’s proof, the following equalities of angles are true.
Complete them with the correct reasoning.
𝛢𝛧̂ 𝛥 = 𝛣𝛢̂ 𝛧 (1)………………………………………………………………………………………………………..
𝛥𝛧̂ 𝛤 = 𝛢𝛣̂ 𝛧 (2)……………………………………………………………………………………………………….
a. (1) as alternate interiorangles
(2) as correspondingangles
b. (1) as correspondingangles
(2) as alternate interiorangles
c. (1) as correspondingangles
(2) as interiorangles
Ερωτήσεις
1. Ποια είναι η επιπλέον γνώση, που μας προσφέρει αυτή η πρόταση, σε σχέση με
προηγούμενες προτάσεις του Ευκλείδη σύμφωνα με τον Πρόκλο;
2. Για ποιο λόγο θεωρεί ο Πρόκλος ότι ήταν αναγκαία η αποδεικτική σειρά των
προτάσεων που ακολούθησε ο Ευκλείδης;
3. Ποιες είναι οι ομοιότητες και οι διαφορές ανάμεσα στην διατύπωση αλλά και στις
αποδείξεις που δίνονται από τον Πρόκλο με τις αντίστοιχες του σχολικού βιβλίου;
4. Ποιες είναι οι ομοιότητες και οι διαφορές ανάμεσα στην διατύπωση αλλά και στις
αποδείξεις που δίνονται από τον Ευκλείδη και τον Πρόκλο;
5. Ποιες είναι οι προτάσεις (θεωρήματα-πορίσματα) πάνω στις οποίες βασίζεται η
αποδεικτική διαδικασία του Πρόκλου για την πρόταση 32;
6. Παρακάτω δίνεται η απόδειξη του Πρόκλου. Δουλεύοντας σε ομάδες, κάντε τις
κατάλληλες συμπληρώσεις ώστε η παρουσίαση της απόδειξης να είναι σύμφωνη
με τα σημερινά δεδομένα.
Προεκτείνουμετηπλευρά ΒΓκαι παίρνουμεστηΒΓ σημείο Ζ.Φέρνουμετην ΑΖκαι απότο Ζ
φέρνουμετηΖΔ παράλληληστηνΑΒ.
𝛢𝛧̂ 𝛥 = 𝛣𝛢̂ 𝛧 (1) ως εντόςεναλλάξ
𝛥𝛧̂ 𝛤 = 𝛢𝛣̂ 𝛧 (2) ως εντόςεκτός
Άρα ολόκληρηηγωνία 𝛢𝛧̂ 𝛤 είναιίση με τις γωνίες 𝛣𝛢̂ 𝛧 , 𝛢𝛣̂ 𝛧
6. Παρομοίως, φέρνουμε την παράλληληκαιαποδεικνύουμεότιηγωνία 𝛢𝛧̂ 𝛣 είναιίση με
τις γωνίες 𝛧𝛢̂ 𝛤 , 𝛢𝛤̂ 𝛧
Άρα οι δύογωνίες 𝛢𝛧̂ 𝛣 και 𝛢𝛧̂ 𝛤 είναιίσες με τις τρειςγωνίες τουτριγώνουΑΒΓ.
Άρα οι τρειςγωνίεςτου τριγώνουΑΒΓείναιίσες με δύο ορθές.
Αλλά καιοι γωνίες 𝛢𝛤̂ 𝛧 και 𝛢𝛤̂ 𝛦 είναιίσες με δύο ορθές. (Γιατί;;)
Ας αφαιρεθεί η κοινή γωνία 𝛢𝛤̂ 𝛧. Άρα η υπόλοιπη εξωτερική γωνία είναι ίση με τις
απέναντι εσωτερικές.
C. You will have to work in groups to do the following tasks.
1. Make a video presentation about Proclus’s proof. Use the video from
task sheet 1 as a model. Your Maths teacher will help you with the
computer programmes that you will need to use.
2. Translate in English the following propositions. Use the glossary in task C
(Task sheet 1) and the two texts given on this Task sheet as a model.
Πρόταση 16
Σε κάθε τρίγωνο , εάν προεκταθεί η μία πλευρά του, η εξωτερική γωνία είναι
μεγαλύτερη καθεμιάς από τις εντός και απέναντι γωνίες.
Πρόταση 17
Σε κάθε τρίγωνο, οι δύο γωνίες είναι μικρότερες των δύο ορθών, με όποιο τρόπο και
αν ληφθούν.
3. Make a presentation(poster,video,PowerPointetc) aboutEuclidandhis
Elements.
7. ΦΥΛΛΟ ΕΡΓΑΣΙΑΣ 3
ΕΝΑ ΑΛΥΤΟ ΠΡΟΒΛΗΜΑ ΤΗΣ ΑΡΧΑΙΟΤΗΤΑΣ
ΤΡΙΧΟΤΟΜΗΣΗ ΟΞΕΙΑΣ ΓΩΝΙΑΣ
Μια ιδέα που βασίζεται στη λύση του Αρχιμήδη
Στοπαρακάτωσχήμα είναι: ΔΒ=ΒΑ=ΑΓ και ΑΔ̂Β= θ
α) Δουλεύονταςσεομάδεςυπολογίστετηγωνία Α̂εξ συναρτήσειτηςγωνίας ΑΔ̂Β
β) Τριχοτομήστετηγωνία Α̂εξ