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 |
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Resultat 1-5 av 5
Side 37
Prove that the opposite sides of the fig . are equal , that the 4 sides = each other ,
3 . that ZS BAD + ADE = 2 right ZS , that the fig . is rectangular ( and : . a square . )
PROPOSITION XLVII . G H Theorem . In any right - ai )gled triangle , the square ...
Prove that the opposite sides of the fig . are equal , that the 4 sides = each other ,
3 . that ZS BAD + ADE = 2 right ZS , that the fig . is rectangular ( and : . a square . )
PROPOSITION XLVII . G H Theorem . In any right - ai )gled triangle , the square ...
Side 39
... a reference to number in this case , viz . , to the number of inches , ( or other
units of length ) in the lines AB , BC . For instance , if Al be 5 inches long , and BC
3 inches , and if on each of the small inch - lines into which AB is divided a
square ...
... a reference to number in this case , viz . , to the number of inches , ( or other
units of length ) in the lines AB , BC . For instance , if Al be 5 inches long , and BC
3 inches , and if on each of the small inch - lines into which AB is divided a
square ...
Side 43
part of it produced , together with the square of half the line bisected , is equal to
the square of the line made up of the half and the part produced . Proved by
showing that cd ' = figs . ch + EF + LG ; and these = AL + CM + LG , which = AM +
LG ...
part of it produced , together with the square of half the line bisected , is equal to
the square of the line made up of the half and the part produced . Proved by
showing that cd ' = figs . ch + EF + LG ; and these = AL + CM + LG , which = AM +
LG ...
Side 44
In the construction of the figure it is necessary to show that BD = GK , in order to
prove that the square BN = square GR , and also to show that BK = KR , in order
to prove that " AK = OM MR . The Demonstration itself consists in showing that AB
...
In the construction of the figure it is necessary to show that BD = GK , in order to
prove that the square BN = square GR , and also to show that BK = KR , in order
to prove that " AK = OM MR . The Demonstration itself consists in showing that AB
...
Side 45
If a straight line be bisected , and produced to any point , the square of the whole
line thus produced , and the square of the part of it produced , are together
double of the square of half the line , and of the square of the line made up of the
half ...
If a straight line be bisected , and produced to any point , the square of the whole
line thus produced , and the square of the part of it produced , are together
double of the square of half the line , and of the square of the line made up of the
half ...
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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 drawn Edition 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 point of contact Problem produced proof PROPOSITION PROPOSITION VIII PROPOSITION XV 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.