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 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.

Let the sum of the interior angles be denoted by I, the number of sides by n, and a right angle by R, then I+4R 2 n R.

COR. 2.-All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal to two right angles (I. 13); therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right 5 angles as there are sides of the figure; that is, by the foregoing corollary, they

A

B

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.

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 (I. 29); and

A

because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC, are equal to

the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD (I. 4), and the triangle ABC to the triangle 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 the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD, equal to one another, AC is parallel to BD (I. 27); and it was shown to be equal to it.

COR. If two equal perpendiculars be drawn to a straight line on the same side of it, the straight line joining their extremities is parallel to the former, and equal to the intercepted part of it.

For the perpendiculars are equal and also parallel (I. 28); therefore the lines joining their extremities are equal and parallel.

The opposite sides and angles of parallelograms are equal to one another, and the diagonal bisects them; that is, divides them in two equal parts.

Let ACDB be a parallelogram, of which BC is a diagonal; the opposite sides and angles of the figure are equal to one another, and the diagonal BC bisects it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal to one another (I. 29); and because AC is parallel to BD,

and BC meets them, the alternate angles ACB, CBD, are equal to one another; wherefore the two triangles ABC, CBD,

B

have two angles ABC, BCA, in one, equal to two angles BCD, CBD, in 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); namely, 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 ACD: And the angle BAC has been shown to be equal to the angle BDC ; there

fore the opposite sides and angles of parallelograms are equal to one another: also, their diagonal 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 the angle BCD; therefore the triangle ABC is equal to the triangle BCD (I. 4), and the diagonal BC divides the parallelogram ACDB into two equal parts. COR. 1.-Parallel lines are equidistant.

For, if from two points in one of them perpendiculars be drawn to the other, they are parallel (I. 28), and the two lines intercepted between them are parallel; therefore a parallelogram is formed, of which the perpendiculars are opposite sides, and are therefore equal.

COR. 2.-Hence two triangles or two parallelograms between the same parallels have the same altitude; and if they have the same altitude, and be on the same side of bases that are in the same straight line, they are between the same parallels. (I. 33, Čor.)

PROPOSITION XXXV. THEOREM.

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

Let the parallelograms ABCD, EBCF (see the 2d and 3d figures), be upon the same base BC, and between the same parallels AF, BC, the parallelogram ABCD shall be equal to the parallelogram EBCF.

D

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 (I. 34); and they 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, because ABCD is a parallelogram, AD is equal to BC; for the same reason EF is equal to BC; wherefore AD is equal to EF (Ax. 1); and DE is common; therefore (Ax. 2 or 3) 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 (I. 29) to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal to the triangle FDC (I. 4);

take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, the remainders therefore are equal (Ax. 3); that is, the parallelogram ABCD is equal to the parallelogram EBCF.

PROPOSITION XXXVI. THEOREM.

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 ÁH, BG, the parallelogram ABCD is equal to EFGH.

Join BE, CH; and because BC is equal to FG, and FG to EH (I. 34),

BC is equal to EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH; therefore (I. 33) EB, CH, are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD (I. 35), because it is upon the same base BC, and between the same parallels BC, AH: For the like reason, the parallelogram EFGH is equal to the same EBCH; therefore also the parallelogram ABCD is equal to EFGH.

PROPOSITION XXXVII. THEOREM.

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

Let the triangles ABC, DBC, be upon the same base BC and between the same parallels AD, BC; the triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B draw BE parallel to CA (I. 31); and E through C draw CF parallel to BD; therefore, each of the figures EBCA, DBCF, is a parallelogram; and EBCA

B

is equal to DBCF (I. 35), 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 diagonal AB bisects it (I. 34); and the triangle DBC is the half of the parallelogram DBCF, because the diagonal DC bisects it. But the halves of equal things are equal (Ax. 7); therefore the triangle ABC is equal to the triangle DBC.

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; the triangle ABC is equal to the triangle DEF.

D

H

Produce AD both ways to the points G, H, and through B draw BG parallel (I. 31) to CA, and through F draw FH parallel to ED. Then each of the figures GBCA, DEFH, is a parallelogram; and they are equal to one another (I. 36), because they are upon

CE

F

equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA (I. 34), because the diagonal AB bisects it; and the triangle DEF is the half of the parallelogram DEFH, because the diagonal DF bisects it. But the halves of equal things are equal (Ax. 7); therefore the triangle ABĈ is equal to the triangle DEF.

PROPOSITION XXXIX. THEOREM.

Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC, be upon the same base BC, and upon the same side of it; they are between the same parallels.

Join AD; AD is parallel to BC; for, if it is not, through the point A draw (I. 31) AE parallel to BC, and join EC; the triangle ABC is equal to the triangle EBC (I. 37), because it is upon the same base BC, and between the same pa