## Euclid's Elements of Geometry: From the Latin Translation of Commandine, to which is Added, a Treatise of the Nature and Arithmetic of Logarithms ; Likewise Another of the Elements of Plane and Spherical Trigonometry ; with a Preface ... |

### Inni boken

Resultat 1-5 av 13

Side 38

THEOR E M. Equal Triangles constituted upon the same Base , on the

, are in the

the

THEOR E M. Equal Triangles constituted upon the same Base , on the

**fame**Side, are in the

**fame**Parallels . ET ABC , DBC , be equal Triangles , constituted uponthe

**fame**Base B C , on the**fame**Side . I say , they are between the same ... Side 139

K + Р Because GH is the

the 1 D same Multiple of AE as GK } H is of A B. But GH is the E + FIMI

Multiple of A E , as L M is of CF. Wherefore GK is G А the

...

K + Р Because GH is the

**fame**Multiple of A E , as HK is of B EB ; therefore GH * isthe 1 D same Multiple of AE as GK } H is of A B. But GH is the E + FIMI

**fame**Multiple of A E , as L M is of CF. Wherefore GK is G А the

**fame**Multiple of A B , as...

Side 147

... as the sixth has to the fourth ; then shall the first , compounded with the fifih ,

have the

has to the fourth ; which was to be demonstrated . PROPOSITION XXV .

PROBLEM .

... as the sixth has to the fourth ; then shall the first , compounded with the fifih ,

have the

**fame**Proportion to the second , as the third compounded with the fixthhas to the fourth ; which was to be demonstrated . PROPOSITION XXV .

PROBLEM .

Side 150

By the same Reason , the

the Triangle ALC be of the Triangle A CD . And if the Bale HC be equal to the

Basc CL , the Triangle A HC is also * equal to the Triangle ALC : And if the Base ...

By the same Reason , the

**fame**Multiple : hat the Base LC is of the Base CD , fhallthe Triangle ALC be of the Triangle A CD . And if the Bale HC be equal to the

Basc CL , the Triangle A HC is also * equal to the Triangle ALC : And if the Base ...

Side 197

THEORE M. if two Right Lines be . perpendicular to one and the

those Right Lines are parallel to one anoiber . LET two Right Lines AB , CD , be

perpendicular to one and the same Plane . I say , A B is parallel to CD . For , let

them ...

THEORE M. if two Right Lines be . perpendicular to one and the

**fame**Plane ,those Right Lines are parallel to one anoiber . LET two Right Lines AB , CD , be

perpendicular to one and the same Plane . I say , A B is parallel to CD . For , let

them ...

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### Vanlige uttrykk og setninger

A B C ABCD added alſo Altitude Bale Baſe becauſe Centre Circle Circumference common Cone conſequently contained Cylinder demonſtrated deſcribed Diameter Difference Diſtance divided double draw drawn equal equal Angles equiangular exceeds fall fame fince firſt fore four fourth given greater half ibis join leſs likewiſe Logarithm Magnitudes manner Multiple Number oppoſite parallel Parallelogram perpendicular Place Plane Point Polygon Priſm produced Prop Proportion PROPOSITION proved Pyramid Quadrant Ratio Reaſon Rectangle remaining Right Angles Right Line Right Line A B ſaid ſame ſay ſecond Segment Series ſhall ſhall be equal Sides ſimilar ſince Sine Solid ſome Sphere Square taken tbis Terms THEOREM thereof theſe third thoſe thro touch Triangle Unity Whence Wherefore whole whoſe Baſe

### Populære avsnitt

Side 193 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.

Side xxiii - If two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to each other; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.

Side 236 - 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 11 - ... sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but...

Side 85 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.

Side 147 - A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.

Side 50 - CB, and to twice the rectangle AC, CB: but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB ; therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line be divided, &c.

Side xxv - EF (Hyp.), the two sides GB, BC are equal to the two sides DE, EF, each to each. And the angle GBC is equal to the angle DEF (Hyp.); Therefore the base GC is equal to the base DF (I.

Side xxxiv - ... therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (26.

Side 194 - ABC, and they are both in the same plane, which is impossible ; therefore the straight line BC is not above the plane in which are BD and BE: wherefore, the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c.