same side BGH, GHD, together equal to two right angles; AB is parallel to CD.

E

Because the angle EGB is equal to GHD, and the angle EGB equal to AGH, the angle AGH is equal to GHD, and they are the alternate angles; therefore AB is parallel to CD. Again, because the angles BGH,GHD, are equal to two right angles; and that AGH, BGH, are equal to two c right angles; the angles AGH, BGH,

A G

H

F

D

are equal to the two BGH, GHD: take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. XXIX. 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 AB, CD; the alternate angles AGH, 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 BGH, GHD, upon the same side, are together equal to two right angles.

A

E

H

B

For, if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater; and because the angle AGH is greater than GHD, add to each of them the angle BGH; therefore the angles AGH, BGH, are greater than BGH, GHD; but the angles AGH, BGH, are equal to two right 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 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 them parallel to EF; AB is also parallel to CD,

A

E

C.

-B

K

-D

Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to GHF. Again, because the straight line GK cuts the parallel straight 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 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.

E

A F

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

B D

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC, equal to one another, EF is parallel to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight 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.

B

I

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

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 ines AC, BD; AC, BD, are also equal and parallel.

A

C

B

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 parallelograms are equal, and the diameter bisects them.

A

B

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 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 equal 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 BC common, 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 between 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 D; it is plain that each of the parallelograms is double of the triangle BDC; and they B are therefore equal to one another.

D

C

F

DE FAE DF

But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal to BC; for the same reason EF is equal to BC;

B

C

B

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 EECF.

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.

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,

B

A

DE

H

CH; therefore EB, CH, are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD; for the like reason, the parallelogram EFGH