## A companion to Euclid: being a help to the understanding and remembering of the first four books. With a set of improved figures, and an original demonstration of the proposition called in Euclid the twelfth axiom, by a graduate |

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Side 13

The student who finds any difficulty in understanding this

very foundation of the science of Geometry , may assist his apprehension of it , by

cutting out the two triangles in paper , and actually applying one to the other ...

The student who finds any difficulty in understanding this

**Theorem**, which is thevery foundation of the science of Geometry , may assist his apprehension of it , by

cutting out the two triangles in paper , and actually applying one to the other ...

Side 14

This is an instance of the argument ad absurdum , i . e . , it is proved that the

and contradiction , and we therefore conclude that it is true . And here let it be

remarked ...

This is an instance of the argument ad absurdum , i . e . , it is proved that the

**theorem**cannot be supposed false , without leading us to a manifest absurdityand contradiction , and we therefore conclude that it is true . And here let it be

remarked ...

Side 21

( Argument ad absurdum ) .

subtended by the greater side , or has the greater side opposite to it . B Steps of

the Demonstration . 1 . Prove that ac cannot be = AB , that ac cannot be < AB ,

that : .

( Argument ad absurdum ) .

**Theorem**. The greater angle of every triangle issubtended by the greater side , or has the greater side opposite to it . B Steps of

the Demonstration . 1 . Prove that ac cannot be = AB , that ac cannot be < AB ,

that : .

Side 33

equal . Steps of the Demonstration . 1 . Prove that OmGc = OM HE , 2 . that the As

are respectively = these equal oms and : . = each other . PROPOSITION XXXIX ...

**Theorem**. Triangles upon the equal bases , and between the same parallels areequal . Steps of the Demonstration . 1 . Prove that OmGc = OM HE , 2 . that the As

are respectively = these equal oms and : . = each other . PROPOSITION XXXIX ...

Side 41

contained by the whole and each of the parts are together equal to the square of

the whole line . Proved by showing that ab % = figures AF + CE ; and that these

are ...

**Theorem**. If a straight line be divided A Ç B into any two parts , the rectanglescontained by the whole and each of the parts are together equal to the square of

the whole line . Proved by showing that ab % = figures AF + CE ; and that these

are ...

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### Vanlige uttrykk og setninger

alternate angle contained angle equal applied Argument ad absurdum base bisect BOOK centre circumference coincides construction Demonstration described diameter directed divided draw drawn Edition Engravings equal equiangular equilateral Euclid extremities fall figure given circle given point given rectilineal given straight line greater HISTORY impossible inscribe interior joins learner least less meet Nature necessary opposite parallel parallelogram pass pentagon point of contact Problem produced proof PROPOSITION PROPOSITION VIII PROPOSITION XV Proved by showing READINGS rectangle contained right angles right line right Zs segment shows the supposition sides similarly square Steps straight line Suppose supposition is false Theorem touch triangle VOLUME whole whole line YOUNG

### Populære avsnitt

Side 24 - 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, namely, either t}le sides adjacent to the equal...

Side 45 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.

Side 18 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

Side 61 - From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle...

Side 37 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Side 76 - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Side 77 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on GEOMETRY.

Side 72 - If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.

Side 27 - If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.