## The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin |

### Inni boken

Side 18

Problem . To

Problem . To

**bisect**a given rectilineal angle , that is , to divide it into two equal angles . E Let BAC be the given rectilineal angle . It is 18 EUCLID'S ELEMENTS . Side 19

It is required to

It is required to

**bisect**it . Construction . In AB take any point from AC cut off AED ... Then the straight line AF shall**bisect**the angle BAC , Proof . ... The angle BAC is**bisected**by the straight line AF . Q.E.F. PROPOSITION 10. Side 22

3 ) ;

3 ) ;

**bisect**FG in H ( I . 10 ) , and join CF , CH , CG . Then the straight line CH drawn from the given point C , shall be perpendicular to the given straight line AB . Proof . Because FH is equal to HG ( constr . ) ... Side 26

**Bisect**AC in E ( I. 10 ) , and join BE ; produce BE to F , making EF equal to BE ( I. 3 ) and join FC . ... In the same manner , if the side BC be**bisected**, and AC be produced to G , it may be demonstrated that the angle BCG , that is ... Side 46

Then the opposite sides and angles of the figure shall be equal to one another , and the diameter BC shall

Then the opposite sides and angles of the figure shall be equal to one another , and the diameter BC shall

**bisect**it . А B D Demonstration . Because AB is parallel to CD , and BC meets them , therefore 1. The angle ABC is equal to the ...### Hva folk mener - Skriv en omtale

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The Elements of Euclid, Containing the First Six Books, with a Selection of ... Euclides Ingen forhåndsvisning tilgjengelig - 2016 |

### Vanlige uttrykk og setninger

ABCD AC is equal alternate angle ABC angle ACB angle BAC base base BC bisected centre circle ABC circumference common compounded constr Construction contained Demonstration describe diameter divided double draw equal angles equiangular equimultiples exterior angle extremities fall figure four fourth given point given straight line greater half inscribed interior join less Let ABC likewise magnitudes manner meet multiple opposite angle parallel parallelogram pass perpendicular plane polygon Problem produced proportionals proved Q.E.D. PROPOSITION ratio reason rectangle rectangle contained rectilineal figure remaining angle right angles segment shown sides similar square square on AC taken third touches the circle triangle ABC unequal wherefore whole

### Populære avsnitt

Side 4 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...

Side 230 - If two triangles, which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line. Let...

Side 110 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Side 207 - ... triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Side 267 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.

Side 197 - 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 21 - 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 61 - If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let the straight line AB be divided into two equal parts...

Side 30 - ... twice as many right angles as the figure has sides ; 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.