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the Right Angle, remaining fixed, the Triangle is turned about till it returns to the Place from whence it first began to move. And if the fixed Right Line be equal to the other that contains the Right Angle, then the Cone is a rectangular Cone; but if it be lefs, it is an obtufe-angled Cone; if greater, an acute-angled Cone. XIX. The Axis of a Cone is that fixed Right Line about which the Triangle is moved.

XX. The Bafe of a Cone is the Circle defcribed by the Right Line moved about.)

XXI. A Cylinder is a Figure defcribed by the Motion of a Right-angled Parallelogram, one of the Sides containing the Right Angle, remaining fixed while the Parallelogram is turned about to the Jame Place from whence it began to be moved. XXII. The Axis of a Cylinder is that fixed Right Line about which the Parallelogram is turned. XXIII. And the Bafes of a Cylinder are the Circles that be defcribed by the Motion of the two oppofite Sides of the Parallelogram.

XXIV. Similar Cones and Cylinders are fuch, whofe Axes and Diameters of their Bafes are proportional.

XXV. A Cube is a folid Figure contained under fix equal Squares.

XXVI. A Tetrahedron is a folid Figure contained under four equal equilateral Triangles.

XXVII. An Octahedron is a folid Figure contained under eight equal equilateral Triangles. XXVIII. A Dodecahedron is a folid Figure contained under twelve equal equilateral and equiangular Pentagons.

XXIX. An Icofabedron is a folid Figure contained under twenty equal equilateral Triangles. XXX. A Parallelepipedon is a Figure contained under fix quadrilateral Figures, whereof thofe which are oppofite are parallel.

PRO

PROPOSITION I.

THE ORE M.

One Part of a Right Line cannot be in a plane Superficies, and another Part above it.

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OR, if poffible, let the Part AB of the Right Line AB C, be in a plane Superficies, and the Part BC above the fame.

There will be fome Right Line in the aforefaid Plane, which with AB will be but one ftrait Line. Let this Line be DB.

Then the two given Right Lines ABC, ABD, have one common Segment AB, which is impoffible; for one Right Line will not meet another in more Points than one. Wherefore, one Part of a Right Line cannot be in a plane Superficies, and another Part above it; which was to be demonftrated.

PROPOSITION II.

THEOREM.

If two Right Lines cut each other, they are both in one Plane, and every Triangle is in one Plane.

ET two Right Lines AB, CD, cut each other

Plane, and every Triangle is in one Plane.

For take any Points, F and G, in the Right Lines AB, CD, and join CB, FG, and let there be drawn FH, GK. In the firft Place, I say, the Triangle EBC is in one Plane.

For if one Part FHC, or GBK, of the Triangle EBC, be in one Plane, and the other Part in anothe: Plane; then one Part of each of the Lines EC, EB, fhall be in one Plane, and the other Part in another 1 of this. Plane; which we have proved to be abfurd. Therefore the Triangle EBC is one Plane, but both the Right Lines EC, E B, are in the fame Plane as the

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Triangle

Triangle BCE is; and AB, CD, are both in the fame Plane as E C, EB are. Wherefore the Right Lines AB, CD, are both in one Plane, and every Triangle is in one Plane; which was to be demonftrated.

PROPOSITION III.

THEOREM.

If two Planes cut each other, their common Section will be a Right Line.

ET two Planes AB, CB, cut each other, whose

is a Right Line.

For if it be not, draw the Right Line DEB in the Plane AB, from the Point D to the Point B, and the Right Line DFB in the Plane BC.

*

Then two Right Lines DEB, DF B, have the fame Terms, and include a Space, which is abfurd. * Axiom 10 Therefore DEB, DF B, are not Right Lines. In the fame manner we demonftrate, that no other Line drawn from the Point D to the Point B, is a Right Line, befides D B, the common Section of the Planes AB, BC. If, therefore, two Planes cut each other, their common Section will be a Right Line; which was to be demonftrated.

PROPOSITION IV.

THEOREM.

If to two Right Lines, cutting one another, a third ftands at Right Angles in the common Section, it fhall be alfo at Right Angles to the Plane drawn thro' the faid Lines.

ET the Right Line EF ftand at Right Angles to

Section E. I fay, EF is alfo at Right Angles to the
Plane drawn thro' AB, CD.

For

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For take the Right Lines EA, EB, CE, DE, equal, and thro' E any how draw the Right Line GEH, and join AD, CB; and from the Point F let there be drawn FA, FG, FD, FC, FH, FB: Then because two Right Lines AE, ED, are equal to two Right Lines CE, EB, and they contain* the equal Angles AED, CEB; the Bafe AD fhall be† equal to the Bafe CB, and the Triangle AED equal to the Triangle CEB; and fo likewife is the Angle DAE equal to the Angle EBC; but the Angle AEG is equal to the Angle BEH; therefore AGE, BEH, are two Triangles, having two Angles of the one equal to the two Angles of the other, each to each; and one Side AE equal to one Side E B, viz. those that are at the equal Angles; and fo the other Sides of the one, will be equal to the other Sides of the other. Therefore GE is equal to EH, and AG to BH; and fince AE is equal to EB, and FE is common and at Right Angles, the Bafe AF fhall be t equal to the Bafe FB: For the fame Reafon likewise, fhall CF be equal to FD. Again, because AD is equal to CB, and AF to FB, the two Sides FA, AD, will be equal to the two Sides FB, BC, each to each; but the Bafe DF has been proved equal to the Bafe FC: Therefore the Angle FAD is equal to the Angle FBC: Moreover, AG has been proved equal to BH; but F B alfo is equal to AF. Therefore the two Sides FA, AG, are equal to the two Sides FB, BH; and the Angle FAG is equal to the Angle F BH, as has been demonftrated; wherefore the Bafe GF is equal to the Bafe FH. Again, becaufe GE has been proved equal to EH, and EF is common, the two Sides GE, EF, are equal to the two Sides HE, EF; but the Bafe HF is equal to the Bafe FG; therefore the Angle GEF is equal to the Angle HEF, and fo both the Angles GEF, HEF, are Right Angles: Therefore FE makes Right Angles with GH, which is any how drawn thro' E. After the fame manner we demonftrate that FE is at Right Angles to all Right Lines that are drawn in *De 3 of the Plane to it; but a Right Line is at Right Angles to a Plane, when it is at Right Angles to all Right Lines drawn to it in the Plane. Therefore FE is at Right Angles to a Plane drawn thro' the Right Lines

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AB, CD. Wherefore, if to two Right Lines cutting one another, a third ftands at Right Angles in the common Section, it shall be alfo at Right Angles to the Plane drawn thro' the faid Lines; which was to be demonftrated.

PROPOSITION V.

THEOREM.

If to three Right Lines, touching one another, a third ftands at Right Angles in their common Seation, thofe three Right Lines fhall be in one and the fame Plane.

LET the Right Line AB ftand at Right Angles

in the Point of Contact B, to the three Right Lines BC, BD, BE. I fay BC, BD, BE, are in one and the fame Plane.

For if they are not, let BD, BE, be in one Plane, and BC above it; and let the Plane paffing thro' AB, BC, be produced, and it will make the common * 3 of this, Section, with the other Plane, a ftrait Line, which let be BF. Then three Right Lines AB, BC, BF, are in one Plane drawn thro' AB, BC; and fince AB ftands at Right Angles to BD and BE, it fhall bet at Right Angles to a Plane drawn thro' BE, † 4 of this DB; and fo AB fhall make ‡ Right Angles with Def. 3. + all Right Lines touching it that are in the fame Plane; but BF being in the faid Plane, touches it. Wherefore the Angle ABF is a Right Angle, but the AngleABC (by the Hyp.) is alfo a Right Angle. Therefore the Angle ABF is equal to the Angle ABC, and they are both in the fame Plane, which cannot be; and fo the Right Line BC is not above the Plane paffing thro' BE and BD. Wherefore the three Lines BC, BD, BE, are in one and the fame Plane. Therefore, if to three Right Lines, touching one another, a third ftands at Right Angles in their common Section, thofe three Right Lines fhall be in one and the fame Planes which was to be demonstrated.

PRO.

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