Πέμπτη 14 Νοεμβρίου 2024
THEOREM OF THE DAY: Euclid’s Triangular Prism
The Elements of Euclid with highlights
Σάββατο 12 Οκτωβρίου 2024
Τετάρτη 9 Οκτωβρίου 2024
Πέμπτη 3 Οκτωβρίου 2024
ΙΣΤΟΡΙΚΟ ΣΗΜΕΙΩΜΑ: Η θεωρία των παραλλήλων | Το αίτημα του Ευκλείδη
Τρίτη 10 Σεπτεμβρίου 2024
Euclid's Elements Book XIII - Proposition 14
Proposition 14
Set out the diameter AB of the given sphere, bisect it at C, describe the semicircle ADB on AB, draw CD from C at right angles to AB, and join DB.
Set out the square EFGH, having each of its sides equal to DB, join HF and EG, set up the straight line KL from the point K at right angles to the plane of the square EFGH, and carry it through to the other side of the plane KM.
Cut off KL and KM from the straight lines KL and KM respectively equal to one of the straight lines EK, FK, GK, or HK, and join LE, LF, LG, LH, ME, MF, MG, and MH.
Then, since KE equals KH, and the angle EKH is right, therefore the square on HE is double the square on EK. Again, since LK equals KE, and the angle LKE is right, therefore the square on EL is double the square on EK.
Κυριακή 11 Αυγούστου 2024
Δευτέρα 22 Ιουλίου 2024
«Ψηφιακός Ευκλείδης: Νέες προοπτικές στη διδασκαλία της Γεωμετρίας»
Κυριακή 30 Ιουνίου 2024
Δευτέρα 3 Ιουνίου 2024
Ψηφιακός Ευκλείδης: Νέες προοπτικές στη διδασκαλία της Γεωμετρίας
Τρίτη 23 Απριλίου 2024
Σάββατο 20 Απριλίου 2024
Euclid's Elements Book XIII - Proposition 13
Proposition 13
Set out the diameter AB of the given sphere, cut it at the point C so that AC is double CB, describe the semicircle ADB on AB, draw CD from the point C at right angles to AB, and join DA.
Set out the circle EFG with radius equal to DC, inscribe the equilateral triangle EFG in the circle EFG, take the center H of the circle, and join EH, HF, and HG.
Set HK up from the point H at right angles to the plane of the circle EFG, cut off HK equal to the straight line AC from HK, and join KE, KF, and KG.
Now, since KH is at right angles to the plane of the circle EFG, therefore it makes right angles with all the straight lines which meet it and are in the plane of the circle EFG. But each of the straight lines HE, HF, and HG meets it, therefore HK is at right angles to each of the straight lines HE, HF, and HG.
And, since AC equals HK, and CD equals HE, and they contain right angles, therefore the base DA equals the base KE. For the same reason each of the straight lines KF and KG also equals DA. Therefore the three straight lines KE, KF, and KG equal one another.