## 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 47

Side 30

Again , because the exterior angle of a triangle is greater than the interior and

opposite angle ( I. 16 ) , therefore 1. The exterior angle BDC of the triangle CDE

is greater than the interior and opposite angle ČED ; for the same

Again , because the exterior angle of a triangle is greater than the interior and

opposite angle ( I. 16 ) , therefore 1. The exterior angle BDC of the triangle CDE

is greater than the interior and opposite angle ČED ; for the same

**reason**, 2. Side 47

... ABCD is a parallelogram , therefore AD is equal to BC ( I. 34 ) ; and for a similar

PROP . XXXIV . 47.

... ABCD is a parallelogram , therefore AD is equal to BC ( I. 34 ) ; and for a similar

**reason**, EF is equal to BC ; wherefore 1. AD is equal to EF ( Ax . 1 BOOK I.-PROP . XXXIV . 47.

Side 49

A ) . And because the parallelograms 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 ) . For the same

A ) . And because the parallelograms 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 ) . For the same

**reason**, 5. Side 56

Euclides James Martin (of the Wedgwood inst, Burslem). Again , because EKHA

is a parallelogram , and AK its diameter , therefore H A B с 2. The triangle AEK is

equal to the triangle AHK ( I. 34 ) ; and for the same

...

Euclides James Martin (of the Wedgwood inst, Burslem). Again , because EKHA

is a parallelogram , and AK its diameter , therefore H A B с 2. The triangle AEK is

equal to the triangle AHK ( I. 34 ) ; and for the same

**reason**, 3. The triangle KGC...

Side 61

For the same

the angle DBC is equal to the angle FBA , each of them being a right angle , add

to each of these equals the angle ABC , therefore 3. The whole angle ABD is ...

For the same

**reason**, 2. BA and AH are in the same straight line . And becausethe angle DBC is equal to the angle FBA , each of them being a right angle , add

to each of these equals the angle ABC , therefore 3. The whole angle ABD is ...

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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.