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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 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 GHD, that is, it is equal to it; but the angle AGH is equal to EGB; therefore likewise EGB is equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH, are equal to BGH, GHD; but EGB, BGH, are equal to two right angles; therefore also BGH, GHD, are equal to two right angles. XXX.-Straight lines which are parallel to the same

straight line are parallel to each other. Let AB, CD, be each of themi parallel to EF; AB 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 A

al to GHF. Again, because the straight Eline GK cuts the parallel straight

C lines EF, CD, the angle GHF is equal to GKD; and it was shown that the angle AGK is equal to GHF; therefore also AGK is equal to GKD; and they are alternate angles; therefore AB is parallel to CD. XXXI.-To draw a straight line through a given point

parallel to a given straight line. Let A be the point, and BC the line; required to draw a straight line through A, parallel to BC.

In BC take point D, join AD, and at A, in AD, make the angle DAE equal to ADC; and produce EA to F.

Because the straight line AD, which B meets the two straight lines BC, EF, makes the alternate angles EAD, ADC, equal to one another, EF is

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parallel to BC. Therefore straight line EAF is drawn through given point A parallel to stratght line BC. XXXII. - 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 BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle, viz., ABC, BCA, CAB, are together equal to two right angles.

Through the point draw CE parallel to AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE, are equal. Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these equals add the angle ACB, and the angles ACD, ACB, are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB, are equal to two right angles: therefore also the angles CBA, BAC, ACB, are equal to two right angles.

Any figure 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. All the angles 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 at the point F, which is the common vertex of the triangles; that is, together with four right angles. Therefore all the angles of the figure, together with fonr right angles. are equal to twice as many right angles as the figure has sides.

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XXXIII.-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 BC; and because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two AB, BC, are equal to the two DC, CB; and the angle ABC is equal to BCD; therefore the base AC is equal to BD, and the triangle ABC to BCD, and the other angles to the other angles each to each, to which the equal sides are opposite ; therefore the angle ACB is equal to CBD; and because the straight line BC meets AC, BD, and makes the alternate angles ACB, CBD, equal to one another, AC is parallel to BD; and it was shown to be equal to it. XXXIV.—The opposite sides and angles of parallelo

grams are equal, and the diameter bisects them. Let ACDB be a parallelogram, of which BC is a diameter. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal to one another; and A ACB, CBD, are equal to one another; wherefore the angles ABC, BCA, are equal to the two angles BCD, CBD, each to each, and one side BC is common; therefore their other sides shall also be equal, each to each, and the third angle of the one to the third angle of the other, viz., the side AB to the side CD, and AC to BD, and the angle BAC equal to BDC; and because the angle ABC is equal to BCD, and CBD to ACB, the whole angle ABD is equal to the whole ACD :and the angle BAC has been shown to be eques

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to BDC; therefore the opposite sides and angles of parallelograms are equal to one another; also, their diameter bisects them; for AB being equal to CD, and BCcommon, the two AB, BC, are equal to the two DC, CB, each to each; and the angle ABC is equal to BCD; therefore the triangle ABC is equal to BCD. XXXV.--Parallelograms upon the same base, and be

tween the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF, be upon the same base BC, and between the same parallels AF, BC.

If the sides AD, DF, of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point I); it is plain that each of the parallelograms is double of the triangle BDC; and they B are therefore equal to one another.

But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, A

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A E DP because ABCD is a parallelogram, AD is equal to BC; for the same reason

B EF is equal to BC; ; wherefore AD is equal to EF; and DE is common; therefore the whole or the remainder AE is equal to the whole or the remainder DF; AB also is equal to DC; and the two EA, AB, are therefore equal to the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB; therefore the base EB is equal to FC, and the triangle EAB equal to FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, then ABCD is equal to EBCF. XXXVI.Parallelograms upon equal bases, and

between the same parallels, are equal to one another. Let ABCD, EFGH, be parallelograms upon equal bases BC, FG, and between the same parallels.

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Join BE, CH; and because BC is equal to FG, and FG to EH, BC is equal to EH; and they are parallels, and joined towards the same parts by BE,

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G CH; therefore EB, CH, are both equal and parallel, and EBCH is a parallelogram; and equals ABCD; for like reason, the parallelogram EFGH equals EBCH; therefore, the parallelogram ABCD equals EFGH. XXXVII.-Triangles upon the same base, and between

the same parallels, are equal to one another. Let the triangles be ABC, DBC.

Produce AD both ways to the ? points E, F, and through B draw BE parallel to CA; and through C draw CF parallel to BD; therefore each of the figures EBCA, DBCF, is a parallelogram; and EBCA is equal to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter A.B bisects it; and DBC is the half of DBCF; therefore ABC is equal to DBC. XXXVIII.— Triangles upon equal bases, and between

the same parallels, are equal to one another. Let the triangles ABC, DEF, be upon equal bases BC, EF, and between the same parallels BF, AD.

Produce AD both ways to the points G, H, and through B draw BG parallel to g CA, and through F draw FH parallel to ED : then each of the figures GBCA, DEFH, is a parallelogram; and they are equal because they are upon equal bases BC, EF, and between the same parallels BF, GH, and ABC is the half of GBCA, because the diameter AB bisects it; and DEF is the half of DEFH, because the diameter DF besects it; but the halves of equals are equal.

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