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" CB, BA, by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC. Then, because the angle at D is a right angle, the angle ACB is greater than a right angle ; (i. "
Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ... - Side 60
av Euclides - 1860
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The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...

Euclid - 1835 - 540 sider
...the rectangle CB, BD, that is, the square of AC alone is less-p. than the squares of CB, BA, by_fc> twice the rectangle CB, BD. Secondly, Let AD fall...triangle ABC: Then, because the angle at D is a right A angle, the angle A CB is greater d than a right angle ; and therefore the square of AB is equal ;...
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...

John Playfair - 1835 - 336 sider
...= AC2 (37. 1.) ; therefore BC' -f AB2=2BC.BD -f AC2 ; that is, AC2 is less than BC2-fAB2 by 2BC.BD. BD Secondly, Let AD fall without the triangle ABC ;* Then because the ani^le at D is a right angle, the angle ACS is greater (9. 1.) than a right angle, and AB2 = (11. 2.)...
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The Element of Geometry

John Playfair - 1836 - 148 sider
...AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD...angle at D is a right angle, the angle ACB is greater (5. 2.) than a right angle ; and therefore the square of AB is equal (26. 2.) to the squares of AC,...
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...

John Playfair - 1836 - 488 sider
...AC2 c ; therefore BC2 + AB2 ij=2 BC.BD + AC2; that is, •» AC2 is less than BC2 + AB2 by 2BC.BD. Secondly, Let AD fall without the triangle ABC * :...the angle at D is a right angle, the angle ACB is greater"1 than a right angle, and AB2 — e AC2 d 16. 1. + BC2 + 2BC.CD. Add BC2 to each ; thenAB2...
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Elements of Plane Geometry According to Euclid

Andrew Bell - 1837 - 290 sider
...; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB-BD. Secondly, let AD fall without the triangle ABC ; then,...angle, the angle ACB is greater than a right angle (1. 16) ; and therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle...
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Elements of Geometry: Containing the First Six Books of Euclid : with a ...

John Playfair - 1837 - 332 sider
...DA2=AC2 (47. 1.) ; therefore BC2+AB2 = 2BC.BD + AC2 ; that is, AC2 is less than BC2+AB2 by 2BC.BD. B nc Secondly, let AD fall without the triangle ABC :*...angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB2= (12. 2.) AC2+BC2+2BC.CD. Add BC2 to each; then AB2+BC'=AC2-|-2BC2+2BC.CD....
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The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...

Robert Simson - 1838 - 434 sider
...that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BB. A Secondly, let AD fall without the triangle ABC : then,...angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle; and therefore the square of AB is equal (12. 2.) to the squares of AC,...
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Euclid's Elements [book 1-6] with corrections, by J.R. Young

Euclides - 1838 - 264 sider
...that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB-BD. , Secondly, let AD fall without the triangle ABC: then,...the angle at D is a right angle, the angle ACB is »j6 1 greater* than a right angle ; and therefore the square of AB ia *It. S. equal* to the squares...
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The Elements of Euclid; viz. the first six books,together with the eleventh ...

Euclides - 1841 - 378 sider
...AC, and twice the rectangle CB, BD; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BD. Secondly, let AD...then, because the angle at D is a right angle, the an* 16. 1. gle ACB is greater* than a right angle; therefore the square of AB * 12. 2. is equal* to...
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...

John Playfair - 1842 - 332 sider
...DA2=AC2 (47. 1.) ; therefore BC2+AB2= 2BC.BD + AC2 ; that is, AC2 is less than BC2+AB2by2BC.BD. BDC Secondly, let AD fall without the triangle ABC :*...angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB2= (12. 2.) AC2+BC2 + 2BC.CD. Add BC2 to each; then AB2+BC2 = A.C2-f2BC2+2BC.CD....
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