## 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. Martin1874 |

### Inni boken

Resultat 1-5 av 33

Side 47

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**ABCD**, EBCF be upon the same base BC , and between the same parallels AF , BC . B Then the parallelogram**ABCD**shall be equal to the parallelogram EBCF . ✓ and therefore E D W B D E F B Demonstration . If the sides AD , DF of the ... Side 48

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**ABCD**is equal to the parallelogram EBCF . Therefore , parallelograms upon the same base , & c . Q.E.D. PROPOSITION 36. - Theorem . Parallelograms upon equal bases , and between the same parallels , are equal to one another . Let**ABCD**... Side 49

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**ABCD**, EBCH are upon the same base BC , and between the same parallels BC , AH ; therefore 4. The parallelogram**ABCD**is equal to the parallelogram EBCH ( I. 35 ) . 5. The parallelogram EFGH is equal to the parallelogram EBCH ; 6. The ... Side 53

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**ABCD**, and the triangle EBC be upon the same base BC , and between the same parallels BC , AE . Then the parallelogram**ABCD**shall be double of the triangle EBC . C Construction . Join AC . B Demonstration . Then D B 1. The triangle ABC ... Side 54

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**ABCD**is double of the triangle ABC , because the diameter AC bisects it ( I. 34 ) ; wherefore also 3.**ABCD**is double of the triangle EBC . Therefore , if a parallelogram and a triangle , & c . Q.E.D. PROPOSITION 42. - Problem . To ...### Andre utgaver - Vis alle

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 Demonstration describe diameter divided double draw equal angles equiangular equimultiples exterior angle extremities fall fore 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 produced Proof proportionals proved Q.E.D. PROPOSITION ratio reason rectangle contained rectilineal figure remaining angle right angles segment shown sides similar square square on AC straight line BC taken third touches the circle triangle ABC unequal wherefore whole

### Populære avsnitt

Side 1 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.

Side 6 - 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 232 - 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 112 - 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 209 - ... 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 269 - 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 199 - 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 23 - 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 63 - 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 32 - ... 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.