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" Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal... "
Elements of Geometry: Containing the First Six Books of Euclid with a ... - Pagina 45
door John Playfair - 1855 - 318 pagina’s
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Euclid in Paragraphs: The Elements of Euclid: Containing the First Six Books ...

Euclid - 1845 - 218 pagina’s
...angle at B is half a right angle, and FDB a right angle, for it is equal* to the interior and•29. :. opposite angle ECB, the remaining angle BFD is half...angle at B is equal to the angle BFD, and the side DF tof the side DB. 1 6. i. And because AC is equal to CE, the square of AC is equal to the square of...
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Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes ...

Euclides - 1845 - 546 pagina’s
...and opposite angle ECB, (i. 29.) therefore the remaining angle BFD is half a right angle ; wherefore the angle at B is equal to the angle BFD, and the side DF equal to the side DB. (i. 6.) And because AC is equal to CE, the square of AC is equal to the square...
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...

Euclid, John Playfair - 1846 - 334 pagina’s
...angle GEF is half a right angle, and EGF a right angle, for it is equal (29. 1.) to the interior and opposite angle ECB, the remaining angle EFG is half...DB. Now, because AC=CE, AC2=CE2, and AC2+CE2=2AC2. BuL(47. 1.) AE2= AC2+CE2 ; therefore AE*=2AC2. Again, because EG=GF, EG2=GF2, and EG2+GF2=2GF2. But...
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The first three books of Euclid's Elements of geometry, with theorems and ...

Euclid, Thomas Tate - 1849 - 120 pagina’s
...Again, because the angle at B is half a right angle and FOB a right angle, for it is equal (i. 29.) to the interior and opposite angle ECB, the remaining...at B is equal to the angle BFD, and the side DF to (i. 6.) the side DB : And because AC is equal to CE, the square of AC is equal to the square of CE;...
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The first two books of the Elements of Euclid, with additional figures ...

Euclides - 1852 - 152 pagina’s
...angle, and FDB a right angle, ° xxlx , i. for it is equal 0 to the interior and opposite angle ECU, the remaining angle BFD is half a right angle; therefore...at B is equal to the angle BFD, and the side DF to f the side DB: And because AC is equal to CE, the square of AC is equal to the square of CE; therefore...
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The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick

Royal Military Academy, Woolwich - 1853 - 400 pagina’s
...Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal (29. i.) to the interior and opposite angle ECB, the remaining...is equal to the angle BFD, and the side DF to (6. i.) the side DB. And because AC is equal to CE, the square of AC is equal to the square of CE ; therefore...
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The first six books of the Elements of Euclid, with numerous exercises

Euclides - 1853 - 178 pagina’s
...right angle, and fdb a right angle, for it is equal (i. 29) to the interior and opposite angle e С b. the remaining angle bfd is half a right angle ; therefore...at b is equal to the angle bfd, and the side df to (i. 6) the side db: and because a С is equal to Сe, the square of a С is equal to the square of...
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The synoptical Euclid; being the first four books of Euclid's Elements of ...

Euclides - 1853 - 146 pagina’s
...Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal (I. 29.) to the interior and opposite angle ECB, the remaining angle BFD is half a right angle ; therefore 7. The angle at B is equal to the angle BFD, and (I. 6.) the side DF is equal to the side DB. And because...
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The Elements of geometry; or, The first six books, with the eleventh and ...

Euclides - 1855 - 262 pagina’s
...interior and opposite angle E С B. Therefore the remaining angle BFD is half a right angle. Wherefore the angle at B is equal to the angle BFD, and the side DF (I. 6) to the side DB. Because AС is equal to СE, the square of AС is equal to the square of СE....
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Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ...

Euclides - 1860 - 288 pagina’s
...(L 6). Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal to the interior and opposite angle ECB, the remaining angle BFD is half a right angle (I. 32) ; therefore the angle at B is equal to the angle BFD, and the side DF to the side DB. Now,...
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