## A Supplement to the Elements of Euclid |

### Inni boken

Resultat 1-5 av 81

Side 6

D is equidistant from A and B ; and is a

the distance DA , which ( constr . ... Let

sides A D I C AB , AC , and CB are bisected in the points D , E and F ,

respectively ...

D is equidistant from A and B ; and is a

**circle**described from D , as a centre , atthe distance DA , which ( constr . ... Let

**ABC**be a given 4 , of which the threesides A D I C AB , AC , and CB are bisected in the points D , E and F ,

respectively ...

Side 8

By the help of this problem a

so as that its circumference shall pass through ... L AGC = _

is equal to the interior opposite L , when the side A C B G BG , of the A AGB · 8 A

...

By the help of this problem a

**circle**may be described about a given triangle ; orso as that its circumference shall pass through ... L AGC = _

**ABC**; i.e. the exterioris equal to the interior opposite L , when the side A C B G BG , of the A AGB · 8 A

...

Side 9

A

obtuse - angled A , obtuse - angled at B , and let ABD be an acute - angled A :

The perpendicular drawn from B to AC falls within AC ; the perpendicular drawn

from ...

A

**circle**cannot cut a straight line in more points than two . ... Let**ABC**be anobtuse - angled A , obtuse - angled at B , and let ABD be an acute - angled A :

The perpendicular drawn from B to AC falls within AC ; the perpendicular drawn

from ...

Side 34

A

triangle as a centre , at the distanće of half the ... Let the point D , in the base BC

of the A

B , of ...

A

**circle**described from the bisection of the ' hypotenuse of a right - angledtriangle as a centre , at the distanće of half the ... Let the point D , in the base BC

of the A

**ABC**, having the B a right angle , be equidistant from either extremity , asB , of ...

Side 78

AH parallel to BC ; from D as a centre , at a distance equal to AF describe a

, cutting AH in G ; join D , G , and through C draw CH parallel to DG : Then is the

ODCHG = A

AH parallel to BC ; from D as a centre , at a distance equal to AF describe a

**circle**, cutting AH in G ; join D , G , and through C draw CH parallel to DG : Then is the

ODCHG = A

**ABC**, and the perimeter of DCHG is , also , equal to the perimeter ...### 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...