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

### From inside the book

Results 1-5 of 57

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**Let**AB be the given finite straight line ; it is required to describe an equilateral triangle upon it . From the ...**ABC**is the equilateral triangle required . Because the point A is the centre of the circle BCD , AC is equal to AB ; ( I ... Page 19

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**Let ABC**, DEF , be two triangles , in which 1. The side AB is equal to the side DE , the side AC equal to the side DF , and the angle A equal to the angle D. It is required to prove that the triangles are equal in every respect ; that ... Page 20

... ABC be applied to DEF , so that the point B shall be on E , and BC on EF , then the point C shall be on F , because ...

... ABC be applied to DEF , so that the point B shall be on E , and BC on EF , then the point C shall be on F , because ...

**Let ABC**be an isosceles triangle , having the side AB equal to the side AC ; it is required to prove that the Α angle ... Page 21

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**let**them stand as in the figure . Then since AB is equal to AC , the angle**ABC**is equal to the angle ACB . But**ABC**is greater than DBC ( ax . 9 ) ; therefore ACB is also greater than DBC , much greater then will the angle DCB be than ... Page 22

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**Let ABC**be a triangle , having the angle ACB equal to the angle ABC ; then shall the side AB , which is opposite the angle ACB , be equal to the side AC , which is opposite the angle ABC . In AB take any point D and make CE equal to BD ...### 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.