The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |
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Side 79
If it be possible , let the two similar segments of circles , viz . ... join CA , DA . and
because the segment ACB is fimilar to the segment A B ADB , and that similar
segments of circles contain equal angles ; the angle ACB is equal to the angle b.
11.
If it be possible , let the two similar segments of circles , viz . ... join CA , DA . and
because the segment ACB is fimilar to the segment A B ADB , and that similar
segments of circles contain equal angles ; the angle ACB is equal to the angle b.
11.
Side 80
Book III . because AB is equal to CD . therefore the straight line AB coinciding
with CD , the segment AEB multe coincide with the segment CFD , and therefore
is equal to it . Wherefore similar sego ments , . & c . l . E. D. 2. 23. 3 . PROP . XXV .
Book III . because AB is equal to CD . therefore the straight line AB coinciding
with CD , the segment AEB multe coincide with the segment CFD , and therefore
is equal to it . Wherefore similar sego ments , . & c . l . E. D. 2. 23. 3 . PROP . XXV .
Side 81
... greater than the angle BAD , the center E falls without the segment ABC , which
therefore is less than a semicircle . but ... greater than a semicircle . wherefore a
segment of a circle being given , the circle is described of which it is a segment .
... greater than the angle BAD , the center E falls without the segment ABC , which
therefore is less than a semicircle . but ... greater than a semicircle . wherefore a
segment of a circle being given , the circle is described of which it is a segment .
Side 85
IN D N a circle , the angle in a semicircle is a right angle ; but the angle in a
segment greater than a semicircle is less than a ... Let ABCD be a circle , of which
the diameter is BC , and center E ; and draw CA dividing the circle into the
segments ...
IN D N a circle , the angle in a semicircle is a right angle ; but the angle in a
segment greater than a semicircle is less than a ... Let ABCD be a circle , of which
the diameter is BC , and center E ; and draw CA dividing the circle into the
segments ...
Side 449
PROP . XCVIII . in jier , P. 2 IF a straight line be drawn - within a circle given in
magnitude cutting off a segment containing a given angle ; if the angle adjacent
to the angle in the segment be bisećied by a straight line produced till it meet the
...
PROP . XCVIII . in jier , P. 2 IF a straight line be drawn - within a circle given in
magnitude cutting off a segment containing a given angle ; if the angle adjacent
to the angle in the segment be bisećied by a straight line produced till it meet the
...
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added alſo altitude angle ABC angle BAC baſe becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole
Populære avsnitt
Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.
Side 24 - ... be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, ECA equal to the angles DEF, EFD, viz.
Side 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Side 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Side 163 - 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 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.
Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.
Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.