Proposition 12
Theorem
In any triangle △ABC , with ∠ACB > 90° , extend BC such that it meets the perpendicular AD from the opposite angle at point D . Then ᅵABᅵ2 = ᅵBCᅵ2 + ᅵCAᅵ2 + 2⋅ᅵBCᅵ⋅ᅵCDᅵ .
ἐν τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ἀμβλϵῖαν γωνίαν ὑποτϵινούσης πλϵυρᾶς τϵτράγωνον μϵῖζόν ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλϵῖαν γωνίαν πϵριϵχουσῶν πλϵυρῶν τϵτραγώνων τῷ πϵριϵχομένῳ δὶς ὑπό τϵ μιᾶς τῶν πϵρὶ τὴν ἀμβλϵῖαν γωνίαν, ἐϕ᾽ ἣν ἡ κάθϵτος πίπτϵι, καὶ τῆς ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλϵίᾳ γωνίᾳ.
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.