## A Supplement to the Elements of Euclid |

### Inni boken

Resultat 1-5 av 47

Side 48

... in the A ABC . Trisect ( S. 33. 1. ) the hypotenuse c , in the points D and E ; from

D and E draw ( E. 11. 1. ) DF and EG 1 to BC , meeting the sides AB and AC in F ,

and G , respectively ; and join F , G : The

... in the A ABC . Trisect ( S. 33. 1. ) the hypotenuse c , in the points D and E ; from

D and E draw ( E. 11. 1. ) DF and EG 1 to BC , meeting the sides AB and AC in F ,

and G , respectively ; and join F , G : The

**inscribed**figure FDEG is a square . Side 52

If any number of parallelograms be

diameters of all the figures shall cut one another in the same point . Let ABCD be

a given , and let EFGH be A E D 7Η K к С І. В G any whatever ,

...

If any number of parallelograms be

**inscribed**in a given parallelogram , thediameters of all the figures shall cut one another in the same point . Let ABCD be

a given , and let EFGH be A E D 7Η K к С І. В G any whatever ,

**inscribed**in ABCD...

Side 55

the 4 AXB and CXD , by the straight line EG , and the LBHC , AXD , by the straight

line FH ; and join E , F , and F , G , and G , H , and H , E : The

EFGH is a square . For , since the figure ABCD is ( hyp . ) equilateral , AC and BD

...

the 4 AXB and CXD , by the straight line EG , and the LBHC , AXD , by the straight

line FH ; and join E , F , and F , G , and G , H , and H , E : The

**inscribed**figureEFGH is a square . For , since the figure ABCD is ( hyp . ) equilateral , AC and BD

...

Side 136

... points of contact will contain a polygon of the same number of angles as the

former ; and any two adjacent angles of the circumscribed figure shall be ,

together , the double of that angle , of the

them .

... points of contact will contain a polygon of the same number of angles as the

former ; and any two adjacent angles of the circumscribed figure shall be ,

together , the double of that angle , of the

**inscribed**figure , which lies betweenthem .

Side 137

... joined : Then it is manifest , that the polygon DCLME has the same number of

angles as AFGHB ; and , further , any two adjacent L A and B of the polygon

AFGHB , are , together , the double of the intermediate L CDE , of the

figure .

... joined : Then it is manifest , that the polygon DCLME has the same number of

angles as AFGHB ; and , further , any two adjacent L A and B of the polygon

AFGHB , are , together , the double of the intermediate L CDE , of the

**inscribed**figure .

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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...