## Elements of Euclid Adapted to Modern Methods in Geometry |

### From inside the book

Results 1-5 of 38

Page 28

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**shewn**in the same manner that the angle BCN is greater than the angle ABC . But BCN is equal to ACD ( I. 11 ) , therefore ACD is greater than ABC ; but it has been before proved to be greater than A ; wherefore the exterior angle ACD is ... Page 31

... ) Again , because CF and FO are greater than CO , ( I. 15. ) if OB be added to both , we shall have CF and FB greater than CO and OB . ( Ax . 4. ) But BA and AC have been

... ) Again , because CF and FO are greater than CO , ( I. 15. ) if OB be added to both , we shall have CF and FB greater than CO and OB . ( Ax . 4. ) But BA and AC have been

**shewn**to be greater BOOK I. 31 PROPS . 15 , 16 . Page 32

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**shewn**to be greater than BF and FC , much more then are BA and AC greater than BO and OC . Again , because BOC is the exterior angle of the triangle OCF , BOC is greater than BFC . ( Ax . 9. ) ( I. 12. ) For the same reason the angle ... Page 37

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**shewn**that they do not meet towards A and C ; therefore they are parallel . Q. E. D. 2. If the angle EFB be equal to the angle FGD . For if the lines AB and CD be not parallel , let them meet in K ; then the exterior angle EFB is ... Page 42

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**shewn**for any number of reflex angles . If the figure be a regular polygon , the magnitude of each exterior angle , compared to a right angle , can easily be found . Cor . 3. - In a right - angled triangle , the two acute angles taken ...### Other editions - View all

Elements of Euclid Adapted to Modern Methods in Geometry Euclid,James Bryce,David Munn (F.R.S.E.) No preview available - 1874 |

### Common terms and phrases

AC and CB altitude angle AOB BA and AC bisecting the angle centre chord circles touch circumference cloth coincide Const conv Cor.-Hence diagonal diameter divided draw equal angles equal to BC equal to twice equiangular equilateral triangle Euclid exterior angle Fcap GEOGRAPHY geometrical given circle given line given point given straight line greater half the perimeter Hence hypotenuse inscribed intersecting isosceles triangle less Let ABC LL.D meet middle point multiple opposite sides parallel to BC parallelogram perpendicular polygon produced Proposition Q. E. D. Cor Q. E. D. PROP radius ratio rectangle contained rectilineal figure reflex angle remaining angles required to prove right angles right-angled triangle schol segments shew shewn side BC square on AC tangent THEOREM triangle ABC twice the rectangle twice the square whole line

### Popular passages

Page 68 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.

Page 77 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Page 50 - If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram is double of the triangle.

Page 87 - 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.

Page 30 - Any two sides of a triangle are together greater than the third side.

Page 204 - Tin; rectangle, contained by the diagonals of a quadrilateral inscribed in a circle, is equal to the sum of the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle...

Page 89 - 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 on the other part.

Page 98 - Guido, with a burnt stick in his hand, demonstrating on the smooth paving-stones of the path, that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.