Euclid, Elements I 48©
translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 48: If the square from one of the sides of a triangle is equal is equal to squares from the remaining two sides of the triangle, the angle enclosed by the remaining sides of the triangle is right.

(general diagram)

Diagrams for book 1 proposition 48, following the display, construction, and demonstration(diagram 1) For in a triangle, ABG, let the square from side BG be equal to the squares from sides BA, AG. I say that the angle by BAG is right.
(diagram 2) For let AD be drawn from point A at right-angles to AG and let AD be positioned equal to BA, and let DG be joined. Since DA is equal to AB, the square from DA is also equal to the square from AB. (diagram 3) Let a common, the square from AG, be added. Therefore, the squares from DA, AG are equal to the squares from BA, AG. (diagram 4) But, that from DG is equal to those from DA, AG. For the angle by DAG is right. But that from BG is equal to those from BA, AG. For it is supposed. Therefore, the square from DG is equal to the square from BG. Thus a side DG is also equal to BG. (diagram 5) And since DA is equal to AB, and AG is common, in fact two, DA, AG, are equal to two BA, AG. And a base, DG, is equal to a base, BG. Therefore, an angle, that by DAG, is equal to an angle, that by BAG. But the angle by DAG is right. Therefore, the angle by BAG is also right. Therefore, if the square from one of the sides of a triangle is equal is equal to squares from the remaining two sides of the triangle, the angle enclosed by the remaining sides of the triangle is right, just what it was required to show.

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