## A Supplement to the Elements of Euclid |

### Inni boken

Resultat 1-5 av 99

Side 5

For , let P be any point , not in DZ , which bisects AB at

possible , let P be equidistant from A and B : Join P , A and P , C ... the ACP =

BCP , and i ' . ... the LACD is a

...

For , let P be any point , not in DZ , which bisects AB at

**right**1 in C ; and , if it bepossible , let P be equidistant from A and B : Join P , A and P , C ... the ACP =

**L**BCP , and i ' . ... the LACD is a

**right**Li .. the < ACP is equal to the 2 ACD , the less...

Side 11

For , if it be possible , let two straight lines meet , which make , with another

straight line , the two interior angles , on the same side , not less than two

Then it is plain , that the three straight lines will thus include a A , two / of which

are ...

For , if it be possible , let two straight lines meet , which make , with another

straight line , the two interior angles , on the same side , not less than two

**right L**:Then it is plain , that the three straight lines will thus include a A , two / of which

are ...

Side 18

Let ACB and EDF be two

also the _DEF =

equal to the perpendicular DG , drawn D to the hypotenuse EF of the A DEF : The

side ...

Let ACB and EDF be two

**right**angled A , AA**right**angled at Ç and D , having ,also the _DEF =

**L**ABC , the 2 EFD =**L**CAB , and the side AC , of the A ABC ,equal to the perpendicular DG , drawn D to the hypotenuse EF of the A DEF : The

side ...

Side 22

2. A rhombus is a parallelogram . PROP . XIX . 29. THEOREM . Every

parallelogram which has one angle a

Let one

angles .

2. A rhombus is a parallelogram . PROP . XIX . 29. THEOREM . Every

parallelogram which has one angle a

**right**angle , has all its angles**right**angles .Let one

**L**, as A , of the ABCD be a**right**angle : The.L B , C , and D are also**right**angles .

Side 23

Let the _XAY be a right 2 : It is required B х D E z Y to trisect it ; i . e . to divide it

into three equal parts . In AX take any point B ; upon AB describe ... draw CE 1 to

AY ; then , since the L BAE , AEC , are

Let the _XAY be a right 2 : It is required B х D E z Y to trisect it ; i . e . to divide it

into three equal parts . In AX take any point B ; upon AB describe ... draw CE 1 to

AY ; then , since the L BAE , AEC , are

**right L**.. ( E. 28. 1. ) AB is parallel to EC ; .### Hva folk mener - Skriv en omtale

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### Vanlige uttrykk og setninger

ABCD aggregate arch base bisect centre chord circle ABC circumference common constr construction describe a circle describe the circle diameter difference distance divided double draw E equal equiangular equilateral extremities fall figure finite straight line four fourth given circle given finite straight given point given ratio given rectilineal given square given straight line greater half inscribed isosceles join less Let ABC lines be drawn locus magnitudes manifest manner mean meet opposite sides parallel to BC parallelogram pass perimeter perpendicular polygon PROBLEM produced PROP proportional proposition rectangle contained rectilineal figure right angles right L scribed segment semi-diameter shewn sides similar square taken tangent THEOREM third touch the circle trapezium triangle vertex vertical whole

### Populære avsnitt

Side 310 - 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 198 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square...

Side 366 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the...

Side 92 - 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.

Side 284 - And if the first have a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth...

Side 349 - Divide a straight line into two parts such that the rectangle contained by the whole line and one of the parts shall be equal to the square on the other part.

Side 288 - Convertendo, by conversion ; when there are four proportionals, and it is inferred, that the first is to its excess above the second, as the third to its excess above the fourth.

Side 296 - ... line and the extremities of the base have the same ratio which the other sides of the triangle have to one...

Side 367 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Side 104 - In every triangle, the square of the side subtending any 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. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular...