the sides of the triangle are bisected.” For let ABΓ be a right-angled triangle, having the angle at B right. And let AB be bisected at Δ. And let ΔE be drawn at right angles. And through E, let EZ be drawn parallel. I say that the remaining sides of the triangle are also bisected, AΓ at E, and BΓ at Z. For let ΔZ be joined. Since AΔ is equal to ΔB, and ΔB to EZ , therefore AΔ is equal and parallel to EZ. And lines joining equal and parallel lines on the same sides are themselves both equal and parallel. But also ΔE, ZΓ. ΓEΔZ is a parallelogram. Therefore ΔZ is equal to EΓ. But it was also equal to AE. Therefore EΓ is equal to AE . Again, since each of BΔEZ, ΓEΔZ is a parallelogram, therefore ΔE is equal to each of BZ, ZΓ; for they are opposite. So that BZ and ZΓ are also equal. The proofs are also for any triangle. Accordingly, then, to these things the width of a river will be measured from a distance. Let there be banks, the one opposite of the enemies on which is point A, and the one towards us ΦH. A split dioptra is fixed in the area towards us, at I, so that the distance from I to the river bank towards us is greater than the river. And this is easy to estimate. And two points are sighted at right angles, one on the opposite bank, either a stone or a bush or some other easily sighted target, and let it be A, and the other point towards us, from the other line of the cross-staff, Y. And moving the dioptra to Y, I sight A and make a right-angled triangle. Let IY be bisected at K. And from K, let KΘ be drawn parallel to AI, and from Θ, let ΘΡ be parallel to IY. Since, therefore, of the right-angled triangle AIY, IY has been bisected at K and ΘK is parallel to AI and ΘΡ to IY , therefore AI is also bisected at P. So measure the distance from I to P. Therefore the distance from P to A is also given. And from this subtracting the distance from P to Φ, we will have the remainder, that is, the width of the river. But if it should seem laborious to anyone to take the greater distance standing back on our own land, it being necessary there for the sight to be disturbed and the result to be confused, we might take the magnitude of the width easily in this manner, standing on the same bank of the river. For again let a point A be taken on the opposite part. And on the part towards us let a point B be taken, so that AB is at right angles to the line along the bank, BΓ. And some point Δ has been taken on BΓ, on which let the rod ΔE be placed. And at the end of the rod let the gnomon at E be raised up, so that if the rod ΔE should touch the surface of the water, the gnomon would be on the surface. And let the rod, at right angles to ΔE , be moved along BΓ, until from some point, Γ , on the line BΓ, the points E and A are seen through a dioptra. And it will be proportional, as BΓ is to ΓΔ, so is AB to EΔ. But the ratio of BΓ to ΓΔ is given. Therefore the ratio of AB to ΔE is also given. And ΔE is given. Therefore AB is also given. By the same reasoning the height of a wall will also be taken, with the same diagram set upright. Let the top of the battlement be A, the base B, and the line from the wall toward us, out of bowshot, be BΓ. A dioptra is hung from a pole (which indeed is called a ‘lampstand’), fixed at right angles at Γ. Let the pole then be the line ΔΓ. Then, inclining the dioptra, I sight the top of the wall, which is A. And having moved to the other vessel, on the same straight line I take a point E. And there will be a triangle AEB, and ΓΔ is parallel to one of the sides, AB. The ratio, therefore, which EΓ has to ΓΔ, the same has EB to BA. And the ratio of EΓ to ΓΔ is given; for each of them is given. Therefore the ratio of EB to AB is also given. And EB is also given , as was shown in the case of the river. Therefore BA is also given, which was to be shown. 1.16 Theft of Sound The Maurousians see more sharply than all and recognize an approaching person from afar and hear equally, although hearing is slower than seeing. But their sight is far-ranging by practice and nature; for in addition to the races in which they train, breathing

τοῦ τριγώνου δίχα τέμνονται πλευραί». Ἔστω γὰρ τρίγωνον ὀρθογώνιον τὸ ΑΒΓ, ὀρθὴν ἔχον τὴν Β γωνίαν. Καὶ τετμήσθω δίχα ἡ ΑΒ, τῷ ∆. Καὶ πρὸς ὀρθὰς ἤχθω ἡ ∆Ε. Καὶ διὰ τοῦ Ε, παράλληλος ἤχθω ἡ ΕΖ. Λέγω ὅτι καὶ <αἱ> λοιπαὶ τοῦ τριγώνου πλευραὶ δίχα τέμνονται, ἡ μὲν ΑΓ κατὰ τὸ Ε, ἡ δὲ ΒΓ κατὰ τὸ Ζ. Ἐπεζεύχθω γὰρ ἡ ∆Ζ. Ἐπεὶ ἴση ἐστὶν ἡ Α∆ τῇ ∆Β, <ἡ δὲ ∆Β τῇ ΕΖ>, ἄρα ἡ Α∆ τῇ ΕΖ ἴση καὶ παράλληλος. Αἱ δὲ ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι ἴσαι τε καὶ παράλληλοί εἰσιν. Ἀλλὰ καὶ αἱ ∆Ε ΖΓ παραλληλόγραμμον τὸ ΓΕ∆Ζ. Ἴση ἄρα ἡ ∆Ζ τῇ ΕΓ. Ἀλλὰ καὶ τῇ ΑΕ ἦν ἴση. <Ἡ ΕΓ ἄρα τῇ ΑΕ ἐστιν ἴση>. Πάλιν, ἐπεὶ ἑκάτερον τῶν Β∆ ΕΖ, ΓΕ ∆Ζ παραλληλόγραμμον, ἡ ∆Ε ἄρα ἴση ἐστὶν ἑκατέρᾳ τῶν ΒΖ ΖΓ· ἀπεναντίον γάρ. Ὥστε καὶ αἱ ΒΖ ΖΓ ἴσαι εἰσίν. Εἰσὶν αἱ ἀποδείξεις καὶ κατὰ παντὸς τριγώνου. Ἀκολούθως δὴ τοῖσδε ποταμοῦ πλάτος ἐκ διαστήματος μετρηθήσεται. Ἔστωσαν ὄχθαι, καταντικρὺ μὲν ἣ τῶν πολεμίων ἐφ' ᾗ σημεῖον τὸ Α, ἣ δὲ πρὸς ἡμᾶς ἡ ΦΗ. Πήγνυται διόπτρα ἐν χώρᾳ τῇ πρὸς ἡμᾶς ἡ σχιστή, κατὰ τὸ Ι, οὕτως ὥστε τὸ διάστημα τὸ τοῦ Ι μέχρι τῆς πρὸς ἡμᾶς ὄχθης τοῦ ποταμοῦ μεῖζον εἶναι τοῦ ποταμοῦ. Τοῦτο δὲ ῥᾴδιον στοχάσασθαι. Καὶ πρὸς ὀρθὰς δύο σημεῖα κατοπτεύεται, ἓν μὲν ἐπὶ τῇ ὄχθῃ καταντικρύ, ἢ λίθος ἢ θάμνος ἤ τις ἄλλος εὐκάτοπτος σκοπός, καὶ ἔστω τὸ Α, τὸ δὲ ἕτερον τὸ πρὸς ἡμᾶς σημεῖον, ἐκ τῆς ἑτέρας τοῦ χιασμοῦ γραμμῆς, τὸ Υ. Τὴν δὲ διόπτραν μεταγαγὼν ἐπὶ τὸ Υ κατοπτεύω τὸ Α καὶ ποιῶ τρίγωνον ὀρθογώνιον. Τετμήσθω ἡ ΙΥ δίχα κατὰ τὸ Κ. Καὶ ἀπὸ τοῦ Κ, τῇ ΑΙ <ἤχθω> παράλληλος ἡ ΚΘ, ἀπὸ δὲ τοῦ Θ, τῇ ΙΥ παράλληλος ἡ ΘΡ. Ἐπεὶ οὖν τριγώνου τοῦ ΑΙΥ ὀρθογωνίου ἡ ΙΥ δίχα τέτμηται τῷ Κ καὶ ἔστι παράλληλος ἡ ΘΚ τῇ ΑΙ <καὶ ἡ ΘΡ τῇ ΙΥ>, καὶ ἡ ΑΙ ἄρα δίχα τέτμηται κατὰ τὸ Ρ. Ἀποτιμᾶν δὴ τὸ ἀπὸ τοῦ Ι ἐπὶ τὸ Ρ διάστημα. ∆έδοται ἄρα καὶ τὸ ἀπὸ τοῦ Ρ ἐπὶ τὸ Α. Τούτου δ' ἀφελόντες τὸ ἀπὸ τοῦ Ρ ἐπὶ τὸ Φ καὶ τὸ λοιπὸν ἕξομεν, τοῦτ' ἔστι τὸ τοῦ ποταμοῦ πλάτος. Εἰ δέ τῳ ἐργῶδες εἶναι δόξει τὸ πλέον ἀποστάντα ἐπὶ τῆς ἡμεδαπῆς διάστημα λαβεῖν, ἀνάγκης ἐκεῖ τούτου γινομένης τὴν ὄψιν ἐπιταράττεσθαι <καὶ> συγχεῖσθαι τὸ γιγνόμενον, λάβοιμεν ἄν, ἐπὶ τῆς αὐτῆς ὄχθης ἑστῶτες τοῦ ποταμοῦ, ῥᾳδίως τὸ μέγεθος τοῦ πλάτους τοῦτον τὸν τρόπον. Ἔστω γὰρ πάλιν ἐπὶ τοῦ καταντικρὺ μέρους εἰλημμένον σημεῖον τὸ Α. Ἐπὶ δὲ τοῦ πρὸς ἡμᾶς μέρους εἰλήφθω σημεῖον τὸ Β, ὥστε εἶναι τὴν ΑΒ πρὸς ὀρθὰς τῇ διὰ τῆς ὄχθης γραμμῇ, τῇ ΒΓ. Εἴληπται δέ τι σημεῖον ἐπὶ τῆς ΒΓ, τὸ ∆, ἐφ' οὗ κανὼν κείσθω ὁ ∆Ε. Ἐπὶ δὲ τοῦ ἄκρου τοῦ κανόνος μετέωρος ἔστω γνώμων ὁ Ε, ὥστε, εἰ ὁ ∆Ε κανὼν τῆς τοῦ ὕδατος ἐπιφανείας ἅπτοιτο, ἐπιπολῆς εἶναι τὸν γνώμονα. Καὶ μέχρι τούτου ὁ κανών, πρὸς ὀρθὰς <τῇ ∆Ε>, τῇ ΒΓ παραφερέσθω, μέχρις οὗ ἀπό τινος <σημείου, τοῦ Γ>, ἐπὶ τῆς ΒΓ γραμμῆς, διὰ διόπτρας θεωρηθῇ σημεῖα τὰ ΕΑ. Καὶ ἔσται ἀνά λογον ὡς ΒΓ πρὸς Γ∆ οὕτως ἡ ΑΒ πρὸς Ε∆. ∆έδοται δὲ ὁ τῆς ΒΓ πρὸς Γ∆ λόγος. ∆έδοται ἄρα καὶ ὁ τῆς ΑΒ πρὸς ∆Ε. Καὶ ἔστιν δοθεῖσα ἡ ∆Ε. ∆οθεῖσα ἄρα καὶ ἡ ΑΒ. Τῷ δὲ αὐτῷ λόγῳ καὶ τείχους ὕψος ληφθήσεται ἐπὶ τοῦ αὐτοῦ διαγράμματος ὀρθουμένου. Ἔστω τὸ μὲν ἄκρον τοῦ προμαχῶνος τὸ Α, βάσις δὲ τὸ Β, ἡ δὲ ἀπὸ τοῦ τείχους εἰς ἡμᾶς ἔξω βέλους γραμμὴ ΒΓ. Κρέμαται διόπτρα ἀπὸ κάμακος (ὃ δὴ «λυχνία» καλεῖται) πηγνυμένη πρὸς ὀρθὰς κατὰ τὸ Γ. Ἔστω δὴ γραμμὴ ὁ κάμαξ ∆Γ. Τὴν δὴ διόπτραν ἐπικλίνας, διοπτεύω τοῦ τείχους τὸ ἄκρον, ὅ ἐστιν Α. Καὶ μετελθὼν ἐπὶ τὸ ἕτερον ἀγγεῖον, ἐπὶ τῆς αὐτῆς εὐθείας λαμβάνω ση μεῖον <τὸ Ε. Καὶ ἔσται τρίγωνον> τὸ ΑΕΒ, καὶ παρὰ μίαν τῶν πλευρῶν τὴν ΑΒ παράλληλος ἡ Γ∆. Ὃν ἄρα λόγον ἔχει ἡ ΕΓ πρὸς Γ∆, τοῦτον ἡ ΕΒ πρὸς ΒΑ. ∆έδοται δὲ ὁ τῆς ΕΓ πρὸς Γ∆ λόγος· δέδοται γὰρ αὐτῶν ἑκατέρα. ∆έδοται ἄρα καὶ ὁ τῆς ΕΒ πρὸς ΑΒ <λόγος. ∆έδοται δὲ καὶ ΕΒ>, ὡς ἐπὶ τοῦ ποταμοῦ δέδεικται. ∆οθεῖσα ἄρα καὶ ἡ ΒΑ, ὅπερ ἔδει δεῖξαι. 1.16 Ἤχου κλοπή Ὀξύτατα πάντων ὁρῶσι Μαυρούσιοι καὶ τὸν προσιόντα γνωρίζουσι μακρόθεν καὶ ἀκούουσιν ἴσα, καίτοι τῆς ἀκοῆς βραδυτέρας οὔσης τοῦ βλέπειν. Ἀλλ' ἡ μὲν ὄψις αὐτοῖς ἀσκήσει καὶ φύσει μακρά· πρὸς γὰρ οἷς γυμνάζονται δρόμοις, ἀναπνέοντες