For let AD be drawn from the point A at right angles to the straight line AC,
let AD be made equal to BA,
and let DC be joined.
Since DA is equal to AB, the square on DA is also equal to the square on AB.
Let the square on AC be added to each;
therefore the squares on DA, AC are equal to the squares on BA, AC.
But the square on DC is equal to the squares on DA, AC,
for the angle DAC is right; [I. 47]
and the square on BC is equal to the squares on BA, AC, for this is the hypothesis;
therefore the square on DC is equal to the square on BC,
so that the side DC is also equal to BC.
And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC;
and the base DC is equal to the base BC;
therefore the angle DAC is equal to the angle BAC. [I. 8]
But the angle DAC is right; therefore the angle BAC is also right.
Ducatur namque B D, perpendicularis ad B A,
& æqualis rectæ B C,
connectaturque recta A D.
Quoniam igitur in triangulo A B D, angulus A B D, rectus est; erit quadratum rectæ A D, æquale quadratis rectarum B A, B D: Est autem quadratum rectæ B D, quadrato rectæ B C, æquale, ob linearum æqualitatem.
Quare quadratum rectæ A D, quadratis rectarum B A, B C, æquale erit.
Cum ergo quadratum rectæ A C, eisdem quadratis rectarum B A, B C, æquale ponatur;
erunt quadrata rectarum A D, A C, inter se æqualia,
ac propterea & rectæ ipsæ A D, A C, æquales.
Quoniam igitur latera B A, B D, trianguli A B D, æqualia sunt lateribus B A, B C, trianguli A B C;
& basis A D, oftensa est æqualis basi A C;
erunt anguli A B D, A B C, æquales:
Est autem angulus A B D, ex constructione rectus. Igitur & angulus A B C, rectus erit.
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Complete text
Book I