falls upon them, the exterior angle ECD is equal (I. 29) to the interior and opposite angle ABC. But the angle ACE was shown to be equal to the angle B AC. Therefore the whole exterior angle A CD is equal (4x. 2) to the two interior and opposite angles CAB, ABC. To each of these equals, add the angle A CB. Therefore the two angles ACD, ACB are equal (Ax. 2) to the three angles C AB, ABC, ACB. But the two angles ACD, ACB are equal (I. 13) to two right angles. Therefore also the three angles CAB, ABC, ACB are equal (Ax. 1) to two right angles. Wherefore, if a side of any triangle be produced, &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.

Let A B C D E be any rectilineal figure. All the interior angles ABC, BCD, &c. together with four right angles are equal to twice as many right angles as the figure has sides.

Divide the rectilineal figure ABCDE 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.

Because the three interior angles of a triangle are equal (I. 32) to two right angles, and there are as many triangles in the figure as it has sides, all the angles of these triangles are equal to twice as many right angles

as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz. A B C, B CD, &c., together with the angles at the point F, which are equal (I. 15. Cor. 2) to four right angles. Therefore all the angles of these triangles are equal (Ax. 1) to the interior angles of the figure together with four right angles. But it has been proved that all the angles of these triangles are equal to twice as many right angles as the figure has sides. 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, made by producing the sides successively in the same direction, are together equal to four right angles.

D

A

B

Because the interior angle ABC, and its adjacent exterior angle ABD, are (I. 13) together equal to two right angles. Therefore all the interior angles, together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with ali the exterior angles are equal (Ax. 1) to all the interior angles and four right angles. Take from these equals all the interior angles. Therefore all the exterior angles of the figure are equal (Ax. 3) to four right angles.

Cor. 1-If two angles of a triangle be given, the third is given; for, it is the difference between their sum and two right angles.

Cor. 2.-If two angles of one triangle be equal to two angles of another triangle, the third angle of the one is equal to the third angle of the other.

Cor. 3.-Every angle of an equilateral triangle is equal to one-third of two right angles, or two-thirds of a right angle. Hence, a right angle can be trisected.

Cor. 4.-If one angle of a triangle be a right angle, the sum of the other two is a right angle.

Cor. 5.-If one angle of a triangle be equal to the sum of the other two, it is a right angle.

Cor. 6. If one angle of a triangle be greater than the sum of the other two, it is obtuse; and if less, acute.

Cor. 7.-In every isosceles right-angled triangle, each of the acute angles is equal to half a right angle.

Cor. 8.-All the interior angles of every quadrilateral figure are together equal to four right angles. This is only a particular case of Euclid's COR. 1; but it is very necessary to be remembered.

Exercise 1.-If a straight line be drawn from one of the angles of a triangle, making the exterior angle equal to the two interior and opposite angles, it is in the same straight line with the adjacent side.

Exercise 2.-To trisect a given finite straight line; that is, to divide it into three equal parts.

Exercise 3.-Any angle of a triangle is right, acute or obtuse, according as the straight line drawn from its vertex bisecting the opposite side, is equal to, greater than, or less than half that side.

Exercise 4.-The straight line drawn from the vertex of any angle of a triangle bisecting the opposite side, is equal to, greater than, or less than half that side, according as the angle is right, acute or obtuse.

Exercise 5.-If the sides of an equilateral and equiangular pentagon or five-sided figure, be produced till they meet, the angles formed at the points of meeting, are together equal to two right angles.

Exercise 6. If the sides of an equilateral and equiangular hexagon, or six-sided figure, be produced till they meet, the angles formed at the points of meeting, are together equal to four right angles.

PROP. 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 the straight lines AC, BD join the two equal and parallel straight lines AB, CD towards the same parts. Then the straight lines AC, BD are also equal and parallel.

B

D

Join B C. Because AB is parallel to CD, and A BC meets them, the angle A B C is equal (I. 29) to the alternate angle BCD. 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, each to each. And the angle ABC was proved to be equal to the angle BCD. Therefore the base AC is equal (I. 4) to the base B D, and the triangle A B C to the triangle BCD. Also, the remaining angles of the one are equal to the remaining angles of the other, each to each; viz., those to which the equal sides are opposite. Therefore the angle ACB is equal to the angle CBD. Because the straight line BC meets the two straight lines A C, BD, and makes the alternate angles ACB, CBD equal to one another.

Therefore AC is (I. 27) parallel to BD; and AC was proved to be equal to BD. Therefore, the straight lines which, &c. Q. Ê. D.

The enunciation of this proposition is more clearly expressed thus: "The straight lines which, without crossing each other, join the extremities of two equal and parallel straight lines, are themselves equal and parallel."

Corollary.-A quadrilateral which has two of its opposite sides equal and parallel, is a parallelogram.--See the following definition :-

DEFINITION XXXVI.

A parallelogram is a four-sided figure of which the opposite sides are parallel; and the diagonal is the straight line joining the vertices of two opposite angles.

PROP. XXXIV. THEOREM.

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.

Let AD 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 B C bisects it.

A

C

B

D

Because AB is parallel to CD, and BC meets them, the angle ABC is equal (I. 29) to the alternate angle BCD. Because AC is parallel to to BD, and BC meets them, the angle ACB is equal (I. 29) to the alternate angle CBD. Because in the two triangles ABC, CBD, the two angles ABC, BCA, in the one, are equal to the two angles B CD, CBD in the other, each to each; and one side BC, adjacent to these equal angles, is common to the two triangles. Therefore their other sides are equal, each to each, and the third angle of the one is equal to the third angle of the other (I. 26); viz., the side AB to the side CD, the side AC to the side BD, and the angle BAC to the angle BDC. Because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB. Therefore the whole angle ABD is equal (Ax. 2) to the whole angle ACD. And the angle BAC has been proved to be equal to the angle BDC. Therefore the opposite sides and angles of a parallelogram are equal to one another.

Also the diagonal BC bisects the parallelogram AD. Because in the two triangles ABC, DCB, AB is equal to CD, and BC common, the two sides AB, BC, are equal to the two sides DC, CB, each to each. And the angle ABC has been proved to be equal to the angle BCD. Therefore the triangle ABC is equal (I. 4) to the triangle BCD. Wherefore the diagonal BC divides the parallelogram AD into two equal parts. Q. E. D.

Corollary 1.-If a parallelogram have one angie a right angle, all its angles are right angles.

Corollary 2.-Parallelograms, having one angle equal in each, are equiangular. Corollary 3.-Parallelograms, which have one angle and two adjacent sides equal, in each, are equal in all respects.

Corollary 4-The adjacent angles of a parallelogram are supplements of each

other.

Exercise 1.-If the opposite sides, or the opposite angles of a quadrilateral figure be equal, it is a parallelogram.

Exercise 2.-The diagonals of a parallelogram bisect each other; and if the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Exercise 3.-The diagonals of rectangular parallelograms are equal; and in oblique-angled parallelograms, those which join the vertices of the acute angles are greater than those which join the obtuse.

Exercise 4.-To divide a straight line into any number of equal parts.

Exercise 5.-To bisect a parallelogram by a straight line drawn through any point in one of its sides.

N.B. In naming a parallelogram by letters, it is customary, and it is sufficient, to take the two letters only which are at the vertices of any two of its opposite angles.

PROP. XXXV. THEOREM.

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

Let the parallelograms A C, BF be upon the same base BC, and and between the same parallels AF, BC. The parallelogram AC is equal to the parallelogram BF.

First, let the sides AD, DF of the parallelograms AC, BF, opposite to the base BC, be terminated in the same point D.

Because each of the parallelograms AC, BF, is double (I. 34) of the triangle BDC. Therefore the parallelogram A C is equal (Ax. 6) to the parallelogram BF.

Next, let the sides AD, EF, opposite to the base BC, be not termirated in the same point.

Because AC is a parallelogram, AD is equal (I. 34) to BC. For a similar reason, EF is equal to BC. Therefore AD is equal (Ax. 1) to EF; and DE is common to both. Wherefore the whole, or the remainder A E, is equal to the whole, or the remainder DF (Ax. 2 or 3); and AB is equal (I. 34) to DC. Because in the triangles EAB, FDC, the side FD is equal to the side EA, and the side DC to the side AB, and the exterior angle FDC is equal (I. 29) to the interior and opposite angle EAB. Therefore the base FC is equal (I. 4) to the base EB, and the triangle FDC _to_the_triangle EAB. From the trapezium ABCF, take the triangle FDC, and the remainder is the parallelogram AC. From the same trapezium take the triangle EAB and the remainder is the parallelogram BF. But when equals are taken from equals, or from the same, the remainders (Ax. 3) are equal. Therefore the parallelogram AC is equal to the parallelogram BF. Therefore, parallelograms upon the same, &c. Q. E.D.

In Dr. Thomson's edition, this highly important proposition is simplified by the application of Prop. XXVI. of this book. It may be simplified still more in the following manner :-Because A B is equal to CD, and BE to CF (I. 34), in the two triangles A BE, DCF', the two sides AB, BE, are equal to the two

sides DC, CF. And the angle ABE is equal to the angle DCF (I. 29 Cor. 1). Therefore, the triangle ABE is equal (I. 4) to the triangle D C F. This equality being proved, the rest of the demonstration is the same as that in the text. That part indeed is often rendered obscure by reference to Axiom 1., instead of a new one, tacitly assumed by Euclid; viz., that "if equals be taken from the same thing, the remainder are equal."

This proposition is the foundation of the mensuration of plane surfaces, and hence of land-measuring. As the area of a rectangle is determined practicaliy by multiplying its length by its breadth, or its base by its altitude, and as by this proposition, every parallelogram having the same base and altitude (that is, the same perpendicular breadth between the parallels) with a rectangle, is equal to that rectangle in area; therefore the area of every parallelogram is found by multiplying the length of its base, by its altitude.

Exercise. Equal parallelograms upon the same base and on the same side of it, are between the same parallels.

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

Let AC, EG be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG. The parallelogram AC is equal to the parallelogram E G.

B

A

DE

C F

H

Join BE, CH. Because BC is equal to FG, (Hyp.) and FG to EH (I. 34), BC is equal to EH (Ax. 1). But BC and EH, are parallel, and joined towards the same parts by the straight lines BE, CH. And straight lines which join the extremities of equal and parallel straight lines towards the same parts, are (I. 33) themselves equal and parallel. Therefore the straight lines BE, CH are both equal and parallel. Wherefore BH is a parallelogram (Def. 36). Because the parallelograms AC, BH, are upon the same base BC, and between the same parallels BC, AH, the parallelogram AC is equal (I. 35) to the parallelogram BH. Because the parallelograms GE, HB are upon the same base EH, and between the same parallels GB, HE, the parallelogram EG is equal to the parallelogram BH. Therefore the parallelogram AC is equal (Ax. 1) to the parallelogram EG. Therefore, parallelograms upon equal bases, &c. Q. Ê. D.

Exercise 1.-If the base of a parallelogram be equal to half the sum of the two parallel sides of a trapezoid, between the same parallels, the parallelogram is equal to the trapezoid.

Exercise 2.-Demonstrate the Theorem of Pappus: The parallelograms described on any two sides of a triangle, are together equal to the parallelogram described on the base, having its side equal and parallel to the straight line drawn from the point of intersection of the exterior sides of the former, to the vertex of the triangle.

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.