## 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 |

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Resultat 1-5 av 36

Side 21

By help of this problem , it may be demonstrated that two straight lines cannot

have a common

two straight lines ABC , ABD . E AB с Construction . From the point B , draw BE at

...

By help of this problem , it may be demonstrated that two straight lines cannot

have a common

**segment**. If it be possible , let the**segment**AB be common to thetwo straight lines ABC , ABD . E AB с Construction . From the point B , draw BE at

...

Side 87

A

circumference which it cuts off . 7 . The angle of a

contained by the straight line and the circumference . 8 . An angle in a

the angle ...

A

**segment**of a circle is the figure contained by a straight line , and thecircumference which it cuts off . 7 . The angle of a

**segment**is that which iscontained by the straight line and the circumference . 8 . An angle in a

**segment**isthe angle ...

Side 88

Similar

contain equal angles . PROPOSITION 1.- Problem . To find the centre of a given

circle . Let ABC be the given circle . It is required to find its centre . с F A B В D E ...

Similar

**segments**of circles are those in which the angles are equal , or whichcontain equal angles . PROPOSITION 1.- Problem . To find the centre of a given

circle . Let ABC be the given circle . It is required to find its centre . с F A B В D E ...

Side 114

PROPOSITION 21. - Theorem . The angles in the same

equal to one another . Let ABCD be a circle , and BAD , BED angles in the same

.

PROPOSITION 21. - Theorem . The angles in the same

**segment**of a circle areequal to one another . Let ABCD be a circle , and BAD , BED angles in the same

**segment**BAED . Then the angles BAD , BED shall be equal to one another . First.

Side 115

Then the

BAC , BEC in the

reason , because CBED is greater than a semicircle , 2. The angles CAD , CED

are equal ...

Then the

**segment**BADC is greater than a semicircle , therefore 1. The anglesBAC , BEC in the

**segment**BADC are equal by the first case . For the samereason , because CBED is greater than a semicircle , 2. The angles CAD , CED

are equal ...

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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.