## 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 9

Side 13

I say , the Right 1 of tbis , Line F C is drawn from the Point C , given in the

common , the two Lines D C , CF , are equal to the two Lines E C , CF ; and the

Base ...

I say , the Right 1 of tbis , Line F C is drawn from the Point C , given in the

**Right****Line A B**, at Right Angles to A B. For , because DC is equal to CE , and F C iscommon , the two Lines D C , CF , are equal to the two Lines E C , CF ; and the

Base ...

Side 15

Wherefore , when a Right Line , standing upon another Right Line , makes

Angles , these shall be either two Right Angles , or ... FOR OR let two Right Lines

B C , BD , drawn from contrary Parts to the Point B , in any

the ...

Wherefore , when a Right Line , standing upon another Right Line , makes

Angles , these shall be either two Right Angles , or ... FOR OR let two Right Lines

B C , BD , drawn from contrary Parts to the Point B , in any

**Right Line A B**, makethe ...

Side 67

ET A B C be a Circle , wherein the Right Line CD , drawn thro ' the Centre ,

bisects the

Angles . For , * find E the Centre of the Circle , ' and let * 1 of this , EA , E B , be

joined .

ET A B C be a Circle , wherein the Right Line CD , drawn thro ' the Centre ,

bisects the

**Right Line A B**, not drawn thro ' the Centre . I say , it cuts it at RightAngles . For , * find E the Centre of the Circle , ' and let * 1 of this , EA , E B , be

joined .

Side 99

For , let the

find F the Centre I of the Cir- | 1 of tbis . cle ; and join EF , FB , FD . Then leke

Angle F E D is . tа

For , let the

**Right Line**D E be drawn * touching * 17 of this . the Circle ABC , andfind F the Centre I of the Cir- | 1 of tbis . cle ; and join EF , FB , FD . Then leke

Angle F E D is . tа

**Right**Angle . And † 18 of this . because D E touches the Circle**AB**... Side 180

LET AB be a Right Line , bifected in the Point C ; and let the Parallelogram A D

be applied to the

and alike fruate to that described on half of the

LET AB be a Right Line , bifected in the Point C ; and let the Parallelogram A D

be applied to the

**Right Line A B**, wanting in Figure the Parailelogram CE , similarand alike fruate to that described on half of the

**Right Line AB**. I say , A D is the ...### Hva folk mener - Skriv en omtale

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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.