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" 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 ...

Euclid, Robert Simson - 1835 - 513 σελίδες
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: 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...

The Elements of Euclid: viz. the first six books, together with the eleventh ...

Euclid, Robert Simson - 1835 - 513 σελίδες
...lines drawn from their centres are equal : Therefore the two sides BG, GC, are equal to the two EH, 'CE HF, and the angle at G is equal to the angle at H ; therefore the base BC is equal" to the base EF: And because the 84. 1. angle at A is equal to the angle at D, the...

The Element of Geometry

John Playfair - 1836 - 114 σελίδες
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences ; the circumference BKC...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...

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

John Playfair - 1836 - 471 σελίδες
...straight lines drawn from their centres are equal : therefore the twq sides BG, GC, are equal to the A two EH, HF ; and the angle at G is equal to the angle a 4. 1. at H ; therefore the base BC is equal * to the base EF : and because the angle at A is equal...

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

John Playfair - 1837 - 318 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 ; 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

Euclid - 1838 - 416 σελίδες
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences : the circumference BKC...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...

The Elements of Euclid; viz. the first six books,together with the eleventh ...

Euclides - 1841
...drawn from their centres » 1 Def. are equal :• therefore the two sides BG, GC are equal •Hyp. to the two EH, HF; and the angle at G is equal* to * 4. 1. the angle at H; therefore the base BC is equal * to the angle at D, the segment BAC is similar...

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

John Playfair - 1842 - 317 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 ; 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...

Euclid in Paragraphs: The Elements of Euclid: Containing the First Six Books ...

Euclid - 1845 - 199 σελίδες
...circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: the circumference BKC is equal to the circumference ELF. Join BC, EF : • Hyp. And because the circles ABC, DEF are equal*, the ti Def. 3. straight lines drawn from their...

Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes ...

Robert Potts - 1845
...are equal: (in. def. I.) therefore the two sides BG, GC, are equal to the two EH, HF, each to each : and the angle at G is equal to the angle at H ; (hyp.) therefore the base BC is equal to the base EF. (i. 4.) And because the angle at A is equal...




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