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 (27. 1.) to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

PROP. XXXII. THEOR.

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 C draw

B

A

C

E

D

CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (29. 1.) 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 shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles 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 (13. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles.

D

COR. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles. 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 at the point F, which is the common vertex of the triangles; that is, (2 Cor. 15. 1.) together with four right angles. Therefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, to

E

A

B

gether with four right angles that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four.

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 (13. 1.) 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.

COR. 3. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

COR. 4. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.

COR. 5. In any triangle there can be but one right angle; for if there were two, the third angle must be nothing. Still less can a triangle have more than one obtuse angle.

COR. 6. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.

COR. 7. Since every equilateral triangle (Cor. 5. 1.) is also equiangular, each of its angles will be equal to the third part of two right angles; so that if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by of one right angle.

COR. 8. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four right angles; hence, if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanctions the Definitions 25 and 26, where the four angles of a quadrilateral are said to be right, in the case of the rectangle and the square.

COR. 9. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles; hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or of one right angle.

COR. 10. The sum of the angles of a hexagon is equal to 2 × (6—2), or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle.

SCHOLIUM.

When (Cor. 1.) is applied to polygons, which have re-entrant angles, as ABC each re-entrant angle must be regarded as greater than two right angles.

And, by joining BD, BE, BF, the figure is divided into four triangles, which contain eight right angles ; that is, as many times two right angles as there are units in the number of sides diminished by two.

But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons with salient angles, which might otherwise be named convex polygons. Every convex polygon is such that a straight line, drawn at pleasure, cannot meet the contour of the polygon in more than two points.

A

PROP. XXXIII. THEOR.

E

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 A

C

B

D

to CD, and BC meets them, the alternate angles ABC, BCD are equal (29. 1.); and 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 (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (4. 1.) 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 (27. 1.) to BD; and it was shewn to be equal to it.

COR. 1. Hence, if two opposite sides of a quadrilateral are equal and parallel, the remaining sides will also be equal and parallel, and the figure will be a parallelogram.

COR. 2. And every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel.

For, having drawn the diagonal BC; then, the triangles ABC, CBD, being mutually equilateral (hyp.), they are also mutually equiangular (Th. 8.), or have their corresponding angles equal; consequently, the opposite sides are parallel; namely, the side AB parallel to CD, and BD parallel to AC; and, therefore, the figure is a parallelogram.

COR. 3. Hence, also, if the opposite angles of a quadrilateral be equal, the opposite sides will likewise be equal and parallel.

For all the angles of the figure being equal to four right angles (Cor. 8.

Th. 32.), and the opposite angles being mutually equal, each pair of adjacent angles must be equal to two right angles; therefore, the opposite sides must be equal and parallel.

PROP. XXXIV. THEOR.

The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it; that is, divides it into two equal parts.

N. B. A Parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is a 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.

Because AB is parallel to CD, and BC A

B

D

meets them, the alternate angles ABC, BCD are equal (29. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (29. 1.) to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and the side BC, which is adjacent to these equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (26. 1.); 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 ACD: And the angle BAC has been shewn to be equal to the angle BDC: therefore the opposite sides and angles of a parallelogram are equal to one another; also, its diameter bisects it; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; now the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (4. 1.) to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts.

COR. 1. Two parallel lines, included between two other parallels, are equal.

COR. 2. Hence, two parallels are every where equally distant. COR. 3. Hence, also, the sum of any two adjacent angles of a parallelogram is equal to two right angles.

PROP. XXXV. THEOR.

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

(SEE THE 20 AND 3d FIGURES.)

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD is equal to the parallelogram EBCF.

If the sides AD, DF of the parallelo- A grams ABCD, DBCF opposite to the base BC be terminated in the same point D; it is plain that each of the parallelograms is double (34. 1.) of the triangle BDC; and they are therefore equal to one another.

B

C

F

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 (34. 1.) to BC; for the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whole, or the remainder, AE is equal (2. or 3. Ax.) to the whole, or the remainder DF; now AB is also equal to DC; therefore the two EA, AB are equal to the two FD, DC, each to each; but the ex

terior angle FDC is equal (29. 1.) to the interior EAB, wherefore the base EB is equal to the base FC, and the triangle EAB (4. 1.) to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders will then be equal (3. Ax.) that is, the parallelogram ABCD is equal to the parallelogram EBCF.

PROP. XXXVI. THEOR.

Parallelograms upon cqual bases, and between the same parallels, are equal to

one another.

Let ABCD, EFGH be parallelograms upon equal bases BC, FG, between the same parallels AH, BG; the parallelogram ABCD

is equal to EFGH.

A

B

DE

C

F

H

and

G

Join BE, CH; and because BC is equal to FG, and FG to (34. 1.) EH, BC is equal to EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH: But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (33. 1.); therefore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal (35. 1.) to ABCD, 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.