## The Elements of Euclid: Viz. the First Six Books, with the Eleventh and Twelfth. In which the Corrections of Dr. Simson are Generally Adopted, But the Errors Overlooked by Him are Corrected, and the Obscurities of His and Other Editions Explained. Also Some of Euclid's Demonstrations are Restored, Others Made Shorter and More General, and Several Useful Propositions are Added. Together with Elements of Plane and Spherical Trigonometry, and a Treatise on Practical Geometry |

### Inni boken

Side 38

See N. TRIANGLES upon the same base , and between the another . Let the

parallels E A D ) F AD , BG : The

Produce AD ...

See N. TRIANGLES upon the same base , and between the another . Let the

**triangles ABC**, DBC be upon the same base BC , and between the fameparallels E A D ) F AD , BG : The

**triangle ABC**is equal to the triangle DBC .Produce AD ...

Side 40

upon equal bases BC , EF , and between the fame parallels BF , AG : But the

equal to the triangle GEF , the greater to the less , which is impossible : Therefore

...

upon equal bases BC , EF , and between the fame parallels BF , AG : But the

**triangle ABC**is B СЕ equal to the triangle DEF ; therefore also the triangle DEF isequal to the triangle GEF , the greater to the less , which is impossible : Therefore

...

Side 41

1 . they are upon equal bases BE , EC , and between the fame parallels BC , AG ;

B C therefore the

parallelogram FECG is likewise double f of the triangle AEC , because it is upon

the fame ...

1 . they are upon equal bases BE , EC , and between the fame parallels BC , AG ;

B C therefore the

**triangle ABC**is double of the triangle AEC : And theparallelogram FECG is likewise double f of the triangle AEC , because it is upon

the fame ...

Side 136

Let the

altitude , viz . the perpendicular drawn from ... BC is to the base CD , so is the

CF.

Let the

**triangles ABC**, ACD , and the parallelograms EC , CF have the samealtitude , viz . the perpendicular drawn from ... BC is to the base CD , so is the

**triangle ABC**to the triangle ACD , and the parallelogram EC to the parallelogramCF.

Side 144

Therefore the angles ABC , DEF are not unequal , that is , they are equal ; and

the angle at A is equal to the angle at D ; therefore the remaining angle at C is

equal to the remaining angle at F : Wherefore the

the ...

Therefore the angles ABC , DEF are not unequal , that is , they are equal ; and

the angle at A is equal to the angle at D ; therefore the remaining angle at C is

equal to the remaining angle at F : Wherefore the

**triangle ABC**is equiangular tothe ...

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### Vanlige uttrykk og setninger

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### Populære avsnitt

Side 30 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.

Side 142 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Side 13 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 30 - ... then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz.

Side 72 - The diameter is the greatest straight line in a circle; and of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the less. Let ABCD be a circle, of which...

Side 57 - If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done: But if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G : and from the centre G, at the distance GB, or GF, describe the semicircle...

Side 145 - AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.

Side 48 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Side 35 - F, which is the common vertex of the triangles ; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 10 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.