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For, if it be not parallel, a b and c d being produced shall meet either towards b, d, or towards a, c: let them be produced and meet towards b, d in the point g; therefore gef is a triangle, and its exterior angle a e fis greater (i. 16) than the interior and opposite angle efg; but it is also

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equal to it, which is impossible; therefore a b and c d being produced do not meet towards b, d. In like manner it may be demonstrated, that they do not meet towards a, c, but those straight lines which meet neither way, though produced ever so far, are parallel (35 def.) to one another. ab therefore is parallel to cd. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION XXVIII.-THEOREM.

If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line, or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.

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LET the straight line ef, which falls upon the two straight lines a b, cd, make the exterior angle egb equal to the interior and opposite angle ghd upon the same side; or b make the interior angles on the same side bgh, ghd together equal to two right angles; a b is parallel to cd.

d

Because the angle egb is equal to the angle ghd, and the angle e gb equal (i. 15) to the angle a gh, the angle agh is equal to the angle ghd ; and they are the alternate angles; therefore a bis parallel (i. 27) to cd. Again, because the angles bgh, ghd are equal (by hyp.) to two right angles; and that a gh, bgh, are also equal (i. 13) to two right angles; the angles a gh, bgh, are equal to the angles bgh, ghd: take away the common angle bgh; therefore the remaining angle a gh is equal to the remaining angle ghd; and they are alternate angles; therefore a b is parallel to cd. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION XXIX.-THEOREM.

If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.

LET the straight line ef fall upon the parallel straight lines a b, cd; the alternate angles a gh, ghd, are equal to one another; and the exterior angle egb is equal to the interior and opposite, upon the same side ghd; and the two interior angles bg h, ghd upon the same side, are together equal to two right angles.

For, if a gh be not equal to ghd, one of them must be greater than the other; let a gh be the greater; and because the angle agh is greater than the angle ghd, add to each of them the angle bgh; therefore the angles a gh, bgh are greater than the angles bg h, ghd; but the angles agh, bgh, are equal (i. 13) to two right

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angles; therefore the angles bgh, ghd, are less than two right angles; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, do meet (12 ax.) together if continually produced; therefore the straight lines ab, cd, if produced far enough, shall meet; but they never meet, since they are parallel by the hypothesis; therefore the angle agh is not unequal to the angle ghd, that is, it is equal to it; but the angle agh is equal (i. 15) to the angle egb; therefore likewise e gb is equal to ghd; add to each of these the angle bgh; therefore the angles eg b, bgh are equal to the angles bgh, ghd; but egb, bgh are equal (i. 13) to two right angles; therefore also bgh, ghd are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION XXX.-THEOREM.

Straight lines which are parallel to the same straight line are parallel to each other.

a

LET a b, c d be each of them parallel to ef; a b is also parallel to cd. Let the straight line ghk cut ab, ef cd; and because ghk cuts the parallel straight lines ab, ef, the angle agh is equal (i. 29) to the angle g hf. Again, because the straight line gk cuts the parallel straight lines ef, cd, the angle ghfis equal (i. 29) to the angle gkd; and it was shewn that the angle agk is equal to the angle ghf; therefore also a gk is equal to gkd; and they are alternate angles; therefore ab is parallel (i. 27) to cd; Wherefore, straight lines, &c. Q. E. D.

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PROPOSITION XXXI.-PROBLEM.

To draw a straight line through a given point parallel to a given straight line.

LET a be the given point, and bc the given straight line; it is required to draw a straight line through the point a, parallel to the straight line bc

b d

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In be take any point d, and join ad; and at the point a, in the straight line a d, make (i. 23) the angle da e equal to the angle ade; and produce the straight line e a to f

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Because the straight line ad, which meets the two straight lines bc, ef, makes the alternate angles e a d, adc equal to one another, ef is parallel (i. 27) to be. Therefore the straight line e af is drawn through the given point a parallel to the given straight line b c. Which was to be done.

PROPOSITION XXXII.-THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

LET abc be a triangle, and let one of its sides be be produced to d; the exterior angle a cd is equal to the two interior and opposite angles cab, a bc, and the three interior angles of the triangle, viz. a bc, bca, cab, are together equal to two right angles. Through the point c draw ce parallel (i. 31) to the straight line a b; and

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because ab is parallel to ce, and ac meets them, the alternate angles bac, a ce, are equal (i. 29). Again, because a b is parallel to ce, and bd falls upon them, the exterior angle e cd is equal to the interior and opposite angle dabe; but the angle ace was shewn to be equal to the angle bac; therefore the whole exterior angle a cd is equal to the two interior and opposite angles ca b, abc; to these equals add the angle a cb, and the angles a cd, a cb are equal to the three angles cba, bac, acb; but the angles a cd, a cb are equal (i. 13) to two right angles: therefore also the angles cba, bac, acb, are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D.

COR. 1. All the interior angles of any rectilineal figure, together with

four right angles, are equal to twice as many right angles as the figure has sides.

For any rectilineal figure abcde, can be divided into as many triangles as the figure has sides, by drawing straight lines from a point f within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure and the same angles are equal to the angles of the figure, together with the angles

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at the point f, which is the common vertex of the triangles: that is (i. 15, cor. 2), together with four right angles. 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.

COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles,

Because every interior angle a bc, with its adjacent exterior a b d, is equal (i. 13) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles.

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PROPOSITION XXXIII.—THEOREM.

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are also themselves equal and parallel.

LET ab, cd be equal and parallel straight lines, and joined towards the same parts by the straight lines ac, bd; ac, bd are also equal and parallel.

Join bo; and because ab is parallel to cd, and b c meets them, the alternate angles abc, bcd are equal (i. 29); and because ab is equal to cd, and bc common to the two triangles a bc, dcb, the two sides a b, bc, are equal to the two dc, cb: and the angle abc is equal to the angle bcd; therefore

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the base ac is equal (i. 4) to the base bd, and the triangle abc to the triangle bcd, and the other angles to the other angles, (i. 4) each to each, to which the equal sides are opposite; therefore the angle a cb is equal to the angle cbd; and because the straight line bc meets the two straight lines a c, bd, and makes the alternate angles acb, cbd equal to one another, a c is parallel (i. 27) to bd; and it was shewn to be equal to it. Therefore straight lines, &c. Q. E. D.

PROPOSITION XXXIV.-THEOREM.

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The opposite sides and angles of parallelograms are equal to one another, the diameter bisects them, that is, divides them into two equal parts. A parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is the straight line joining two of its opposite angles.

Let acdb be a parallelogram, of which bc is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter bc bisects it.

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Because a b is parallel to cd, and bc meets them, the alternate angles

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abc, bcd are equal (i. 29) to one another and because a c is parallel to bd, and b c meets them, the alternate angles a cb, cbd, are equal (i. 29) to one another; wherefore the two triangles abc, cbd d have two angles a bc, bca in one, equal to two angles bcd, cb din the other, each to each, and one side bc common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other (i. 26), viz. the side ab to the side cd, and ac to bd, and the angle bac equal to the angle bdc and because the angle abc is equal to the angle bcd, and the angle cbd to the angle acb, the whole angle abd is equal to the whole angle a cd: and the angle bac has been shewn to be equal to the angle bdc; therefore the opposite sides and angles of parallelograms are equal to one another; also, their diameter bisects them; for a b being equal to cd, and bc common, the two ab, bc are equal to the two dc, cb, each to each; and the angle a bc is equal to the angle bcd; therefore the triangle a bc is equal (i. 4) to the triangle bcd, and the diameter bc divides the parallelogram ac db into two equal parts. Q. E. D.

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PROPOSITION XXXV.-THEOREM.

Parallelograms upon the same base, and between the same parallels, are equal to one another.

LET the parallelograms a b c d, ebcf (see the second figure) be upon the same base bc, and between the same parallels af, bc; the parallelogram abcd shall be equal to the parallelogram e bcf.

If the sides ad, df of the parallelograms abcd, dbcf, opposite to the base bc, be terminated in the same point d; it is plain that each of

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the parallelograms is double (i. 34) of the triangle bdc; and they are therefore equal to one another.

But, if the sides ad, ef (see second and third figures), opposite to the base bc of the parallelograms a b c debcf, be not terminated in the same point d; then, because abcd is a parallelogram, ad is equal (i. 34) to bc; for the same reason ef is equal to

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