§1## α΄. Πᾶς κύκλος ἴσος ἐστὶ τριγώνῳ ὀρθογωνίῳ, οὗ ἡ μὲν ἐκ τοῦ κέντρου ἴση μιᾷ τῶν περὶ τὴν ὀρθήν, ἡ δὲ περίμετρος τῇ βάσει.
## Proposition 1 Any circle is equal to a right-angled triangle in which the radius is equal to one of the sides about the right angle, and the perimeter is equal to the base.
Ἐχέτω ὁ ΑΒΓ△ κύκλος τριγώνῳ τῷ Ε, ὡς ὑπόκειται· λέγω ὅτι ἴσος ἐστίν.
Let the circle ΑΒΓΔ have [this relation] to the triangle Ε, as assumed; I say that it is equal.
Εἰ γὰρ δυνατόν, ἔστω μείζων ὁ κύκλος, καὶ ἐγγεγράφθω τὸ ΑΓ τετράγωνον, καὶ τετμήσθωσαν αἱ περιφέρειαι δίχα, καὶ ἔστω τὰ τμήματα ἤδη ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κύκλος τοῦ τριγώνου· τὸ εὐθύγραμμον ἄρα ἔτι τοῦ τριγώνου ἐστὶ μεῖζον.
For, if possible, let the circle be greater, and let the square ΑΓ be inscribed, and let the arcs be bisected, and let the segments be already less than the excess by which the circle exceeds the triangle; therefore the rectilinear figure is still greater than the triangle.
Εἰλήφθω κέντρον τὸ Ν καὶ κάθετος ἡ ΝΞ· ἐλάσσων ἄρα ἡ ΝΞ τῆς τοῦ τριγώνου πλευρᾶς.
Let the center Ν and the perpendicular ΝΞ be taken; therefore ΝΞ is less than the side of the triangle.
Ἔστιν δὲ καὶ ἡ περίμετρος τοῦ εὐθυγράμμου τῆς λοιπῆς ἐλάττων, ἐπεὶ καὶ τῆς τοῦ κύκλου περιμέτρου ἔλαττον ἄρα τὸ εὐθύγραμμον τοῦ Ε τριγώνου· ὅπερ ἄτοπον.
And the perimeter of the rectilinear figure is also less than the remaining [side], since it is also less than the perimeter of the circle; therefore the rectilinear figure is less than the triangle Ε; which is absurd.
Ἔστω δὲ ὁ κύκλος, εἰ δυνατόν, ἐλάσσων τοῦ Ε τριγώνου, καὶ περιγεγράφθω τὸ τετράγωνον, καὶ τετμήσθωσαν αἱ περιφέρειαι δίχα, καὶ ἤχθωσαν ἐφαπτόμεναι διὰ τῶν σημείων· ὀρθὴ ἄρα ἡ ὑπὸ ΟΑΡ. Ἡ ΟΡ ἄρα τῆς ΜΡ ἐστὶν μείζων·
But let the circle be, if possible, less than the triangle Ε, and let the square be circumscribed, and let the arcs be bisected, and let tangents be drawn through the points; therefore the angle ΟΑΡ is a right angle.
ἡ γὰρ ΡΜ τῇ ΡΑ ἴση ἐστί· καὶ τὸ ΡΟΠ τρίγωνον ἄρα τοῦ ΟΖΑΜ σχήματος μεῖζόν ἐστιν ἢ τὸ ἥμισυ.
Therefore ΟΡ is greater than ΜΡ; for ΡΜ is equal to ΡΑ; and the triangle ΡΟΠ is also greater than half of the figure ΟΖΑΜ.
Λελείφθωσαν οἱ τῷ ΠΖΑ τομεῖ ὅμοιοι ἐλάσσους τῆς ὑπεροχῆς, ᾗ ὑπερέχει τὸ Ε τοῦ ΑΒΓ△ κύκλου· ἔτι ἄρα τὸ περιγεγραμμένον εὐθύγραμμον τοῦ Ε ἐστὶν ἔλασσον· ὅπερ ἄτοπον· ἔστιν γὰρ μεῖζον, ὅτι ἡ μὲν ΝΑ ἴση ἐστὶ τῇ καθέτῳ τοῦ τριγώνου, ἡ δὲ περίμετρος μείζων ἐστὶ τῆς βάσεως τοῦ τριγώνου.
Let the remaining [figures] similar to the sector ΠΖΑ be left less than the excess by which the triangle Ε exceeds the circle ΑΒΓΔ; therefore the circumscribed rectilinear figure is still less than the triangle Ε; which is absurd; for it is greater, because ΝΑ is equal to the perpendicular of the triangle, and the perimeter is greater than the base of the triangle.
Ἴσος ἄρα ὁ κύκλος τῷ Ε τριγώνῳ.
Therefore the circle is equal to the triangle Ε.
§2## β΄. Ὁ κύκλος πρὸς τὸ ἀπὸ τῆς διαμέτρου τετράγωνον λόγον ἔχει, ὃν ῑᾱ πρὸς ῑδ.
## Proposition 2 The circle has to the square on its diameter the ratio which 11 has to 14.
Ἔστω κύκλος, οὗ διάμετρος ἡ ΑΒ, καὶ περιγεγράφθω τετράγωνον τὸ ΓΗ, καὶ τῆς Γ△ διπλῆ ἡ △Ε, ἕβδομον δὲ ἡ ΕΖ τῆς Γ△.
Let there be a circle whose diameter is ΑΒ, and let the square ΓΗ be circumscribed about it, and let ΔΕ be double of ΓΔ, and ΕΖ be one-seventh of ΓΔ.
Ἐπεὶ οὖν τὸ ΑΓΕ πρὸς τὸ ΑΓ△ λόγον ἔχει, ὃν κᾱ πρὸς ζ, πρὸς δὲ τὸ ΑΕΖ τὸ ΑΓ△ λόγον ἔχει, ὃν ἑπτὰ πρὸς ἐν, τὸ ΑΓΖ πρὸς τὸ ΑΓ△ ἐστίν, ὡς κβ πρὸς ζ.
Since, then, the triangle ΑΓΕ has to the triangle ΑΓΔ the ratio which 21 has to 7, and the triangle ΑΓΔ has to the triangle ΑΕΖ the ratio which 7 has to 1, the quadrilateral ΑΓΔΖ is to the triangle ΑΓΔ as 22 is to 7.
Ἀλλὰ τοῦ ΑΓ△ τετραπλάσιόν ἐστι τὸ ΓΗ τετράγωνον, τὸ δὲ ΑΓ△Ζ τρίγωνον τῷ ΑΒ κύκλῳ ἴσον ἐστίν·
But the square ΓΗ is four times the triangle ΑΓΔ, while the figure ΑΓΔΖ is equal to the circle ΑΒ.
ὁ κύκλος οὖν πρὸς τὸ ΓΗ τετράγωνον λόγον ἔχει, ὃν ῑᾱ πρὸς ιδ.
Therefore, the circle has to the square ΓΗ the ratio which 11 has to 14.