Bibliotheca Polyglotta

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.

IN amblygoniis triangulis, quadratum, quod fit a latere angulum obtusum subtendente, maius est quadratis, quæ fiunt a lateribus obtusum angulum comprehendentibus, rectangulo bis comprehenso & ab vno laterum, quæ sunt circa obtusum angulum, in quod, cum protractum fuerit, cadit perpendicularis, & ab assumpta exterius linea sub perpendiculari prope angulum obtusum.

三邊鈍角形之對鈍角邊上直角方形。大於餘邊上兩直角方形幷之較。為鈍角旁任用一邊、偕其引增線之與對角所下垂線相遇者、矩內直角形。二。

Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced; I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD.

For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4] Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD. But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47] therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.

Therefore etc. Q. E. D.

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