Therefore AC is (I. 27) parallel to BD; and AC was proved to be equal to B D. 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.

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

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 BCD, 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 AC, BF be upon the same base BC, and and between the same parallels AF, BC. The parallelogram AC is equal to the parallelogram B F.

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 EA B. 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 ABE, 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 DCF. 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 practically 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

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. È. 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.

Produce AD both ways to the points E and E F. Through B draw BE parallel to CA (I. 31), and through C draw CF parallel to BD. Then each of the figures E C, BF, is a parallelogram (Def. 36.)

The parallelograms EC, BF, are equal (I. 35), because they are upon the same base

BC, and between the same parallels BC, EF. The triangle ABC is half of the parallelogram EC (I. 34), because the diagonal AB bisects it. Also, the triangle DBC is half of the parallelogram B F, 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 D B C. Wherefore, triangles, &c. Q. E. D.

Exercise. To describe a triangle equal to any given rectilineal figure.

In solving this problem, the learner may begin with a parallelogram; then proceed to a trapezium, a pentagon, a hexagon, and so on. He will then ultimately find that, if a figure had a thousand sides, he could reduce it, by degrees, to an equivalent triangle.

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.

G

A

D

H

B

C E

Produce AD both ways to the points G and H. Through B draw BG parallel to CA (I. 31), and through F draw FH parallel to ED. Then each of the figures GC, EH, is a parallelogram. The parallelograms GC, EH are equal to one another (Î. 36), because they are upon equal bases BC, EF, and between the same parallels BF, GH. The triangle ABC is the half of the parallelogram GC (I. 34), because the diagonal A B bisects it. And the triangle DEF is the half of the parallelogram EH, because the diagonal DF bisects it. But the halves of equal things are equal (Ax. 7). Therefore the triangle ABC is equal to the triangle DE F. Wherefore, triangles upon equal bas es, &c. Q. E. D.

Corollary 1.-The straight line drawn from any angle of a triangle bisecting the opposite side, bisects the triangle.

Corollary 2.-If two triangles have two sides of the one equal to two sides of the other, each to each, and the angle contained by the two sides of the one, the supplement of the angle contained by the two sides of the other, these triangles are equal.

Exercise 1.-To bisect a triangle by drawing a straight line through any point in one of its sides.

Exercise 2.-The straight lines drawn from the opposite angles of a parallelogram to any point in a diagonal, divide it into equal triangles on opposite sides of the segments of the diagonal.

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 A same base BC, and upon the same side of it. The triangles ABC, DBC, are between the same parallels; that is, if AB be joined, AD is parallel to BC.

B

For if AD be not parallel to DC, through the point A draw A E parallel to BC (I. 31), meeting BD in E; and join EC.

The triangle A B C is equal to the triangle E B C (I. 37), because they are upon the same base B C, and between the same parallels B C, A E. But the triangle ABC is equal (Hyp.) to the triangle DBC. Therefore the triangle DBC is equal to the triangle EB C, the greater to the less, which is impossible. Therefore AE is not parallel to B C. In the same manner it can be proved that no other straight line but AD is parallel to BC. Therefore AD is parallel to B C. Wherefore, equal triangles upon, &c. Q. E. D.

Exercise 1.-If two sides of a triangle be bisected, the straight line joining the points of bisection is parallel to the base, and equal to half of it.

Exercise 2.-If the sides of any quadrilateral figure be bisected, and the points of bisection be joined, the figure thus formed will be a parallelogram, and equal to half the quadrilateral.

Equal triangles upon equal bases in the same straight line, and on the same side of it, are between the same parallels.

Let the equal triangles AB C, DEF, be upon equal bases B C, EF, in the same straight line BF, and on the same side of it; they are between the same parallels; that is, if AD be joined, AD is parallel to BF.

For, if AD be not parallel to B F, through A draw AG parallel to BF (I. 31), meeting ED in G, and join GF.

A

D

The triangle ABC is equal to the triangle GEF (I. 38), because they are upon equal bases BC, EF, and between the same parallels BF, A G. But the triangle ABC is equal (Hyp.) to the triangle DEF. Therefore the triangle DEF is equal (Ax. 1) to the triangle GEF, the greater equal to the less, which is impossible. Therefore A G is not parallel to BF. In the same manner it can be proved that no other straight line is parallel to BF, but AD. Therefore AD is parallel to B F. Wherefore, equal triangles upon, &c. Q. E. D. In this proposition and the preceding one, the demonstration would be conducted in the same manner, if the parallels A E and A G were to meet BD and ED produced. The argument would then rest on the absurdity of the less triangle being equal to the greater.

D