## A companion to Euclid: being a help to the understanding and remembering of the first four books. With a set of improved figures, and an original demonstration of the proposition called in Euclid the twelfth axiom, by a graduate |

### Inni boken

Side 80

Prove that similarly DG = DF ; and : : DE , DF , DG = each other , 3 . that a o

described with centre d , and distance either of these lines , will touch the sides

AB , BC , CA , and : : that O EFG is

describe a ...

Prove that similarly DG = DF ; and : : DE , DF , DG = each other , 3 . that a o

described with centre d , and distance either of these lines , will touch the sides

AB , BC , CA , and : : that O EFG is

**inscribed**in A ABC . Proposition V . Todescribe a ...

Side 81

To

. 1 . Prove that ( in As ABE , AED ) base AB = base ad , that similarly BC , CD =

BA , or AD ; and : : AB , BC , CD , DA = each other , that _ Bad is a right Z , that the

...

To

**inscribe**a square bk . . . . . in a given circle . aiman Steps of the Demonstration. 1 . Prove that ( in As ABE , AED ) base AB = base ad , that similarly BC , CD =

BA , or AD ; and : : AB , BC , CD , DA = each other , that _ Bad is a right Z , that the

...

Side 82

To

that the opposite sides of figs . AK , KP , AH , HD , AG , GC , BG , GD = each other

, that AE = AF , and . . FG = GE , that GE , GF , GH , GK = each other , that if a o ...

To

**inscribe**a circle in a given square . H Steps of the Demonstration . aj 1 . Provethat the opposite sides of figs . AK , KP , AH , HD , AG , GC , BG , GD = each other

, that AE = AF , and . . FG = GE , that GE , GF , GH , GK = each other , that if a o ...

Side 84

To

of the Demonstration . 1 . Prove that 5 ZS DAC , ACE , ECD , CDB , and BDA =

each other , that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE ...

To

**inscribe**an equilateral and equiangular pentagon in a given circle . ai ni Stepsof the Demonstration . 1 . Prove that 5 ZS DAC , ACE , ECD , CDB , and BDA =

each other , that AB , BC , CD , DE , EA = each other , that the pentagon ABCDE ...

Side 87

To

of the Demonstration . 1 . Prove that A EGD is equilateral , that / EGD = of 2 right <

s , and similarly < DGC = f of 2 right ZS , that < CGB = { of 2 right ZS , and : : S ...

To

**inscribe**an equilateral and equiangular hexagon in a given circle . oi o Stepsof the Demonstration . 1 . Prove that A EGD is equilateral , that / EGD = of 2 right <

s , and similarly < DGC = f of 2 right ZS , that < CGB = { of 2 right ZS , and : : S ...

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### Vanlige uttrykk og setninger

alternate angle contained angle equal applied Argument ad absurdum base bisect BOOK centre circumference coincides construction Demonstration described diameter directed divided draw Edited Engravings equal equiangular equilateral Euclid extremities fall figure given circle given point given rectilineal given straight line greater HISTORY impossible inscribe interior joins learner least less meet Nature necessary opposite parallel parallelogram pass pentagon Problem produced proof PROPOSITION Prove Prove that ZS Proved by showing READINGS rectangle contained right angles right line right Zs segment shows the supposition sides similarly square Steps straight line Suppose supposition is false Theorem touch triangle VOLUME whole whole line Young

### Populære avsnitt

Side 24 - If two triangles have two angles of the [one equal to two angles of the other, each to each, and one side equal to one side, namely, either t}le sides adjacent to the equal...

Side 45 - 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 of the other part.

Side 18 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

Side 61 - From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle...

Side 37 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Side 76 - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Side 77 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on GEOMETRY.

Side 72 - If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.

Side 27 - If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.