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Βιβλία Βιβλία 11 - 20 από 51 για ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines....
" ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is... "
The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The ... - Σελίδα 81
των Robert Simson - 1804
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The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...

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...equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumterences : The circumference BKC is equal to the circumference ELF. Join BC, EF ; and because the circles ABC, DEF, are equal, the straight lines drawn from their centres are equal : Therefore...

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John Playfair - 1819 - 317 σελίδες
...ABC, DEF are equal, the straight lines drawn from their centres are equal : therefore the two A sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4. 1.) to the base EF : and because the angle at A is equal to the angle at D, the...

The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...

Euclid, Robert Simson - 1821 - 516 σελίδες
...ABC, DEF are equal, the straight lines drawn from their centres are equal; therefore the two sides BG, GC are equal to the two EH, HF; and the angle at G is equal to the angle at H;thereforethe base BC is equal (4. 1. Jto the base EF; and because the angle at A is equal to the...

A Popular Course of Pure and Mixed Mathematics ...: With Tables of ...

Peter Nicholson - 1825 - 372 σελίδες
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF et their centres, and ВАС, EDF at their circumferences : The circumference BKC...equal to the circumference ELF. Join BC, EF ; and because the circles ABC, DEF, are equal, the »raight lin?s drawn from their centres are equal : Therefore...

Elements of arithmetic, algebra and geometry

George Lees - 1826 - 207 σελίδες
...Then, because the circles are equal, their radii are equal ; therefore, the two sides BG, GC, are B equal to the two EH, HF ; and the angle at G is equal to the angle at Hs ; therefore, the base BC is equal to the base EFa. Conceive the circle ABC applied to the circle...

Elements of Geometry: Containing the First Six Books of Euclid, with a ...

Euclid, John Playfair - 1826 - 320 σελίδες
...straight lines drawn from their eentres are equal: theref«re tJie tw« A / 43^ nr mf ""•' (a iides BG, GC, are equal to the two EH, HF; and the angle at G is equ .Ito the angle at H; therefore the base BC is equal (4. i.) to the base EF : and beeavse the angle...

Elements of Geometry, Containing the First Six Books of Euclid

Euclid, Phillips - 1826 - 180 σελίδες
...^S\\ ) ( / /\\ ) • Def. I. lines drawn from their centres shall also be equal : therefore the two BG, GC, are equal to the two EH, HF, and the angle at G equal to the angle at н. Wherefore also the base Bcb is equal to the base EF. Again, because the angle...

Euclid's Elements of geometry, transl. To which are added ..., Βιβλία 1-6

Euclides - 1826
....Д\ ) ( / _Д\ } • Def. I. lines drawn from their centres shall also be equal: therefore the two BG, GC, are equal to the two EH, HF, and the angle at G equal to the angle at н. Wherefore also the base at A is equal to the angle at D, the segment BAC...

The Elements of Euclid: Viz, the First Six Books, Together with the Eleventh ...

Euclid, Robert Simson - 1829 - 516 σελίδες
...drawn from their centres are equal : therefore the two sides BG, GC are equal to the two EH, HF ; and K the angle at G is equal to 'the angle at H ; therefore the base BC is equal (4. 1. ) to the base EF; and because the angle at A is equal to the angle atD, the...

Elements of Geometry: Containing the First Six Books of Euclid : with a ...

John Playfair - 1832 - 333 σελίδες
...are equal, the .straight lines drawn from their centres are equal : therefore the two AD CE sides EG, GC, are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal (4. 1.) to the base EF: and because the angle at A is equal to the angle at D, the...




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