§11.prop.11ἀπὸ τοῦ δοθέντος σημείου μετεώρου ἐπὶ τὸ δοθὲν ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.
To draw a straight line perpendicular to a given plane from a given elevated point.
ἔστω τὸ μὲν δοθὲν σημεῖον μετέωρον τὸ Α, τὸ δὲ δοθὲν ἐπίπεδον τὸ ὑποκείμενον· δεῖ δὴ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.
Let the given elevated point be A, and the given plane the plane set below; it is required to draw a straight line perpendicular to the plane set below from the point A.
διήχθω γάρ τις ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖα, ὡς ἔτυχεν, ἡ ΒΓ, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ. εἰ μὲν οὖν ἡ ΑΔ κάθετός ἐστι καὶ ἐπὶ τὸ ὑποκείμενον ἐπίπεδον, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν.
For let some straight line BC be drawn at random in the plane set below, and let AD be drawn from the point A perpendicular to BC. If indeed AD is also perpendicular to the plane set below, that which was prescribed would be done.
εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΒΓ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΔΕ, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΔΕ κάθετος ἡ ΑΖ, καὶ διὰ τοῦ Ζ σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΗΘ.
καὶ ἐπεὶ ἡ ΒΓ ἑκατέρᾳ τῶν ΔΑ, ΔΕ πρὸς ὀρθάς ἐστιν, ἡ ΒΓ ἄρα καὶ τῷ διὰ τῶν ΕΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
But if not, let DE be drawn from the point D at right angles to BC in the plane set below, and let AZ be drawn from A perpendicular to DE, and let HT be drawn through the point Z parallel to BC. And since BC is at right angles to each of the straight lines DA, DE, therefore BC is also at right angles to the plane through EDA.
καί ἐστιν αὐτῇ παράλληλος ἡ ΗΘ· ἐὰν δὲ ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· καὶ ἡ ΗΘ ἄρα τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
And HT is parallel to it; but if there be two parallel straight lines, and one of them is at right angles to any plane, the remaining one will also be at right angles to the same plane; therefore HT is also at right angles to the plane through ED, DA.
καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΗΘ. ἅπτεται δὲ αὐτῆς ἡ ΑΖ οὖσα ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ· ἡ ΗΘ ἄρα ὀρθή ἐστι πρὸς τὴν ΖΑ·
And HT is therefore at right angles to all the straight lines meeting it and being in the plane through ED, DA. And AZ, being in the plane through ED, DA, meets it; therefore HT is at right angles to ZA; so that ZA is also at right angles to TH.
ὥστε καὶ ἡ ΖΑ ὀρθή ἐστι πρὸς τὴν ΘΗ. ἔστι δὲ ἡ ΑΖ καὶ πρὸς τὴν ΔΕ ὀρθή· ἡ ΑΖ ἄρα πρὸς ἑκατέραν τῶν ΗΘ, ΔΕ ὀρθή ἐστιν.
But AZ is also at right angles to DE; therefore AZ is at right angles to each of the straight lines HT, DE.
ἐὰν δὲ εὐθεῖα δυσὶν εὐθείαις τεμνούσαις ἀλλήλας ἐπὶ τῆς τομῆς πρὸς ὀρθὰς ἐπισταθῇ, καὶ τῷ διʼ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ἡ ΖΑ ἄρα τῷ διὰ τῶν ΕΔ, ΗΘ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
But if a straight line be set up at right angles to two straight lines cutting one another at their intersection, it will also be at right angles to the plane through them; therefore ZA is also at right angles to the plane through ED, HT.
τὸ δὲ διὰ τῶν ΕΔ, ΗΘ ἐπίπεδόν ἐστι τὸ ὑποκείμενον· ἡ ΑΖ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
But the plane through ED, HT is the plane set below; therefore AZ is at right angles to the plane set below.
ἀπὸ τοῦ ἄρα δοθέντος σημείου μετεώρου τοῦ Α ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος εὐθεῖα γραμμὴ ἦκται ἡ ΑΖ· ὅπερ ἔδει ποιῆσαι.
Therefore, from the given elevated point A, the straight line AZ has been drawn perpendicular to the plane set below; which was to be done.
§11.prop.12τῷ δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ δοθέντος σημείου πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι.
To erect a straight line at right angles to a given plane from a given point on it.
ἔστω τὸ μὲν δοθὲν ἐπίπεδον τὸ ὑποκείμενον, τὸ δὲ πρὸς αὐτῷ σημεῖον τὸ Α· δεῖ δὴ ἀπὸ τοῦ Α σημείου τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι.
Let the given plane be the plane set below, and the point on it A; it is required to erect a straight line at right angles to the plane set below from the point A.
νενοήσθω τι σημεῖον μετέωρον τὸ Β, καὶ ἀπὸ τοῦ Β ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος ἤχθω ἡ ΒΓ, καὶ διὰ τοῦ Α σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΑΔ.
ἐπεὶ οὖν δύο εὐθεῖαι παράλληλοί εἰσιν αἱ ΑΔ, ΓΒ, ἡ δὲ μία αὐτῶν ἡ ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ ἡ λοιπὴ ἄρα ἡ ΑΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
Let some elevated point B be conceived, and from B let the perpendicular BG be drawn to the plane set below, and through the point A let AD be drawn parallel to BG. Since indeed two straight lines AD, GB are parallel, and one of them BG is at right angles to the plane set below, therefore the remaining one AD is also at right angles to the plane set below.
τῷ ἄρα δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ σημείου τοῦ Α πρὸς ὀρθὰς ἀνέσταται ἡ ΑΔ· ὅπερ ἔδει ποιῆσαι.
Therefore, to the given plane, from the point A on it, AD has been erected at right angles; which was to be done.
§11.prop.13ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς οὐκ ἀναστήσονται ἐπὶ τὰ αὐτὰ μέρη.
From the same point, two straight lines cannot be erected at right angles to the same plane on the same side.
εἰ γὰρ δυνατόν, ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Α τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι αἱ ΑΒ, ΑΓ πρὸς ὀρθὰς ἀνεστάτωσαν ἐπὶ τὰ αὐτὰ μέρη, καὶ διήχθω τὸ διὰ τῶν ΒΑ, ΑΓ ἐπίπεδον· τομὴν δὴ ποιήσει διὰ τοῦ Α ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν.
For, if possible, from the same point A let two straight lines AB, AC be erected at right angles to the plane set below on the same side, and let the plane through BA, AC be drawn; indeed it will make a straight line as intersection through A in the plane set below.
ποιείτω τὴν ΔΑΕ· αἱ ἄρα ΑΒ, ΑΓ, ΔΑΕ εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ.
Let it make DAE; therefore the straight lines AB, AC, DAE are in one plane.
καὶ ἐπεὶ ἡ ΓΑ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας.
And since GA is at right angles to the plane set below, therefore it will make right angles with all the straight lines meeting it and being in the plane set below.
ἅπτεται δὲ αὐτῆς ἡ ΔΑΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ἡ ἄρα ὑπὸ ΓΑΕ γωνία ὀρθή ἐστιν.
And DAE, being in the plane set below, meets it; therefore the angle GAE is right.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΕ ὀρθή ἐστιν· ἴση ἄρα ἡ ὑπὸ ΓΑΕ τῇ ὑπὸ ΒΑΕ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ·
For the same reason indeed, BAE is also right; therefore the angle GAE is equal to the angle BAE.
ὅπερ ἐστὶν ἀδύνατον.
And they are in one plane; which is impossible.
οὐκ ἄρα ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς ἀνασταθήσονται ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἔδει δεῖξαι.
Therefore, from the same point, two straight lines cannot be erected at right angles to the same plane on the same side; which was to be proved.