## Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement of the Quadrature of the Circle and the Geometry of SolidsF. Nichols, 1806 - 311 sider |

### Fra bogen

Resultater 1-5 af 55

Side 36

... , AD is equala to BC ; for the same

... , AD is equala to BC ; for the same

**reason**EF is equal to BC : where- fore AD is equal to EF ; and DE is common ; therefore the whole , or the remainder AE is equal to the whole , or the remainder DF ; and AB is also equal to 36 ELEMENTS. Side 42

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**reason**the triangle KGC is equal to the triangle KFC . Then , because the triangle AEK is equal to the triangle AHK , and the triangle KGC to the tri- angle KFC , the triangle AEK , together with the triangle KGC , are equal to the ... Side 46

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**reason**, AB and AH are in the same straight line . Now because the angle DBC is d 14. 1 . e 10. Ax . equale to the angle FBA , each of them being a right 2. Ax . g 4. 1 . h 41. 1 . k 6. Ax . angle , add to each the an- B D G H L E K gle ... Side 53

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**reason**HF also is a square , and it is upon the side HG , which is equal to AC ; therefore HF , CK are the squares of AC , CB . The complement AG is equal to the h 43. 1 . complement GE , and AG - AC.CG - AC.CB ; therefore GE = AC.CB ... Side 56

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**reason**PR is equal to RO ; and because CB is equal to BD , and GK to KN , the rectangles CK and BN are equal , as also the rectangles GR and RN . But CK is equal to RN , because they are the complements of the parallelogram CO ...### Andre udgaver - Se alle

### Almindelige termer og sætninger

ABC is equal ABCD altitude angle ABC angle ACB angle BAC angle contained angle EDF arch base BC bisected Book centre circle ABC circumference coincide cylinder definition demonstrated diameter draw drawa equal angles equiangular equilateral polygon equimultiples Euclid exterior angle fore four right angles given circle given straight line greater inscribed interior and opposite join less Let ABC Let the straight meet multiple opposite angle parallelogram perpendicular point F polygon prism PROB produced proportional proposition pyramid Q. E. D. COR Q. E. D. PROP ratio rectangle contained rectilineal figure remaining angle segment solid angle solid parallelepipeds straight line AB straight line AC Suppl THEOR third touches the circle triangle ABC triangle DEF

### Populære passager

Side 121 - 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. either the sides adjacent to the equal...

Side 42 - TO a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Side 63 - Therefore, in obtuse-angled triangles, &c. QED PROP. XIII. THEOREM. In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.

Side 3 - 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 183 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG ; the ratio of the parallelogram AC to the parallelogram CF is the same with the ratio which is compounded •f the ratios of their sides.

Side 3 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.

Side 291 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 160 - ... extremities of the base shall have the same ratio which the other sides of the triangle have to one...

Side 10 - ... shall be greater than the base of the other. Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz.

Side 14 - Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extretnity equal to one another.