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XXIX. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.

A. A parallelopiped is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel.

B

PROP. I. THEOR, One part of a straight line cannot be in a plane, and

another part without it. If it be possible, let AB, part of the straight line ABC, be in a plane, and the part BC without it : Then, since the straight line AB is in the plane, it can be produced in that plane (1. Ax. 2): Let it be produced to D; and let any plane pass through the straight line AD, and be turned about it until it pass through the point C: Now, because the points B, C are in this plane, the straight line BC is in it (1. Def. 7): Therefore the two straight lines ABC, ABD are in the same plane and have a common segment AB—which (1. 11. Cor.) is impossible: Therefore AB and BC are in the same plane.

Wherefore, One part of a straight line &c. Q.E.D.

PROP. II. THEOR, Two straight lines which cut one another are in one plane;

and three straight lines which meet one another are in one plane.

Let the two straight lines AB, CD cut one another in E: AB and CD shall be in one plane; and the three

B

straight lines EC, CB, BE, which meet one another, are in one plane.

Let any plane pass through EB, and be turned about EB, produced if necessary, until it A pass through the point C: Then, because the points E, C are in this plane, the straight line EC is in it. In like manner, the straight line BC is in the same plane: And (Hyp.) EB is also in it; therefore the three straight lines EC, CB, BE are in one plane: But in the plane in which EC, BE are, in the same are CD, AB (11. 1); therefore AB and CD are in one plane.

Wherefore, Two straight lines &c. Q. E, D.

PROP. III. THEOR. If two planes cut one another, their common section is a

straight line. Let two planes AB, BC cut one another, and let the line BD be their common section: BD shall be a straight line.

For, if not, from the point B to D, draw, in the plane AB, the straight line BED, and in the plane BC, the straight line BFD : Then the two straight lines BED, BFD have the same extremities, and therefore include a space betwixt themwhich (1. Ax. 10) is impossible: Therefore BD, the common section of the planes AB, BC, cannot but be a straight line.

Wherefore, If two planes &c.

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D

А C

Q, E.D.

PROP. IV. THEOR. If a straight line stand at right angles to each of two

straight lines in the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are.

Let the straight line EF stand at right angles to each of the two straight lines AB, CD, in E, the point of their intersection : EF shall also be at right angles to the plane passing through AB, CD.

Take AE, EB, CE, ED, all equal to one another ; join AD, CB, and through E draw, in the plane in which are AB, CD, any straight line cutting AD, CB in G and H; and from any point F, in EF, draw FA, FG,FD, FC, FH, FB: Then, because the two AE, ED are equal to the two BE, EC, each to each, and that they contain equal angles AED, BEC— D therefore the base AD is equal to the base BC, and the angle DAE to the angle EBC: And the angle AEG is equal to the angle BEH; therefore the triangles AEG, BEH have two angles of the one equal to two angles of the other, each to each, and the sides AE, EB, adjacent to the equal angles, equal to one anothertherefore GE is equal to EH, and AG to BH: And because AE is equal to EB, and EF common and at right angles to them--therefore the base AF is equal to the base BF; and, in like manner, CF is equal to DF: And because the two sides DA, AF are equal to the two CB, BF, each to each, and the base DF to the base CF-therefore the angle DAF is equal to the

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angle CBF: Again, because the two GA, AF are equal to the two HB, BF, each to each, and that they contain equal angles—therefore the base GF is equal to the base HF: And, lastly, because the two GE, EF are equal to the two HE, EF, each to each, and the base GF to the base HF-therefore the angle GEF is equal to the angle HEF, and consequently each of these angles is a right angle : Therefore EF makes right angles with GH, that is, with any straight line drawn through E, in the plane passing through AB, CD: Therefore (11. Def. 3) EF is at right angles to the plane in which are AB, CD.

Wherefore, If a straight line &c.

Q.E.D.

PROP. V. THEOR. If three straight lines meet all in one point, and a straight

line stand at right angles to each of them in that point, the three straight lines shall be in one and the same plane,

Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet : BC, BD, BE shall be in one and the same plane.

For, if not, let, if possible, BD and BE be in one plane, and BC without it; and let a plane pass through AB and BC, the common section of which, with the plane in which are BD and BE, will be a straight line-BF suppose: Then the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB and BC:

And because AB stands at right angles to each of the straight

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lines BD, BE, it is also at right angles to the plane passing through them (11. 4): Therefore it makes right angles with every straight line which meets it in that plane, and therefore the angle ABF is a right angle: But the angle ABC (Hyp) is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane-which is impossible : Therefore the straight line BC is not without the plane in which are BD and BE, that is, the three straight lines BC, BD, BE are in one and the same plane.

Wherefore, If three straight lines, &c.

Q.E.D.

PROP. VI. THEOR. If tro straight lines be at right angles to the same plane,

they shall be parallel to one another. Let the straight lines AB, CD be at right angles to the same plane: AB shall be parallel to CD.

Let them meet the plane in the points B, D, and join BD, to which draw DE at right angles, in the same plane; make DE equal to AB, and join BE, AE, AD: Then, because AB is perpendicular to the plane, it makes right angles with every straight line which meets it in that plane : But BD, BE, which are in that plane, do each of them meet AB; therefore each of the angles ABD, ABE is a right angle: In like manner, each of the angles CDB, CDE is a right angle: And because the two AB, BD are equal to the two ED, DB, each to each, and that they contain right angles—therefore the base AD is equal to the base EB : Again, because the two AB, BE are equal to the two

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