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alſo angle B A becauſe the angle biſected caſe centre circle A B circumference circumſcribed cone conſequent cylinder D E F demonſtrated deſcribed diameter equal angles equiangular equimultiples Euclid fides firſt given circle given right line given triangle greater inſcribed join leſs leſſer magnitudes oppoſite P R O B parallel parallelogram paſſing thro perpendicular point F polygon priſms prop proportional propoſition pyramid reŽtangle remaining angle right angles right line A B right lined figure right-lined ſaid ſame altitude ſame multiple ſame ratio ſame reaſon ſay ſecond ſegment ſemidiameter ſhall ſides A B ſimilar ſince ſolid angle ſome ſphere ſtand ſuch ſum ſuperficies T H E O theſe thoſe trapezium triangle A B twice the ſquare uſe vertex the point Wherefore whoſe baſe
Pagina 247 - 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.
Pagina 248 - But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But by the hypothesis, it is less than a right angle ; which is absurd.
Pagina 18 - When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
Pagina 32 - Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.
Pagina 56 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Pagina 391 - KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB to...
Pagina 110 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Pagina 130 - When you have proved that the three angles of every triangle are equal to two right angles...
Pagina 183 - FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of...