angles, and the parallelogram a rectangle. (See cors. to th. 16.)

Corol. 2. Hence, also, the sum of any two adjacent angles of a parallelogram is equal to two right angles. Corol. 3. If two parallelograms have an angle in each equal, the parallelograms are equiangular.

THEOREM XX.

A.

B

Every quadrilateral whose opposite sides are equal is a parallelogram, or has its opposite sides parallel. Let ABDC be a quadrangle having the opposite sides equal, namely, the side AC equal to BD, and AB equal to CD; then shall these equal sides be also parallel, and the figure a parallelogram.

D

For, let the diagonal BC be drawn. Then the triangles ABC, CBD being mutually equilateral (by hyp.), they are also mutually equiangular (th. 5), or have their corresponding angles equal; consequently, the opposite sides, having the same difference of direction in opposite ways from the same line BC, have the same direction one way, and are parallel (def. 8); viz., the side AB parallel to DC, and AC parallel to BD, and the figure is a parallelogram (def. 30). Q. E. D.

THEOREM XXI.

Those lines which join the corresponding extremities of two equal and parallel lines are themselves equal and parallel.

Let AB, DC be two equal and parallel lines; then will the lines AC, BD, which join their extremes, be also equal and parallel. [See the fig. above.]

For, draw the diagonal BC. Then, because AB and DC are parallel (by hyp.), the angle ABC is equal to the alternate angle DCB (th. 10). Hence, then, the two triangles having two sides and the con

tained angles equal, viz., the side AB equal to the side DC, and the side BC common, and the contained angle ABC equal to the contained angle DCB, they have the remaining sides and angles also respectively equal (th. 1); consequently AC is equal to BD, and also parallel to it (th. 11). Q. E. D.

General Scholium. From the foregoing theorems it appears that a quadrilateral will be a parallelogram: 1. When it has its opposite sides parallel; 2. When it has its opposite sides equal; 3. When it has two of its opposite sides equal and parallel. A quadrilateral is also a parallelogram: 4. When two sides are parallel and two opposite angles equal; 5. When the opposite angles are equal. The proof of the last two cases is left as an exercise for the learner.

THEOREM XXII.

Parallelograms, as also triangles, standing on the same base, and between the same parallels, are equal to each other.

same base AB, and between the same parallels AB, DE; then will the parallelogram ABCD be equal to the parallelogram ABEF, and the triangle ABC equal to the triangle ABF.

For the two triangles ADF, BCE are equiangular, having their corresponding sides in the same direction; and having the two corresponding sides AD, BC equal (th. 19), being opposite sides of a parallelogram, these two triangles are identical, or equal in all respects (th. 2). If each of these equal triangles, then, be taken from the whole space ABED, there will remain the parallelogram ABEF in the one case,

equal to the parallelogram ABCD in the other (by ax. 3).

Also, the triangles ABC, ABF, on the same base AB, and between the same parallels, are equal, being the halves of the said equal parallelograms (th. 19).* Q. E. D.

Corol. 1. Parallelograms, or triangles, having the same base and altitude are equal. For the altitude is the same as the perpendicular or distance between the two parallels, which is every where equal, by theorem 12.

Corol. 2. Parallelograms, or triangles, having equal bases and altitudes are equal. For, if the one figure be applied with its base on the other, the bases will coincide, or be the same, because they are equal; and so the two figures, having the same base and altitude, are equal.

THEOREM XXIII.

If a parallelogram and a triangle stand on the same base, and between the same parallels, the parallelogram will be double the triangle, or the triangle half the parallelogram.

Let ABCD be a parallelogram, and D ABE a triangle, on the same base AB, and between the same parallels AB, DE; then will the parallelogram ABCD be double the triangle ABE, or the triangle half the parallelogram.

A

C E

B

For, draw the diagonal AC of the parallelogram, dividing it into two equal parts (th. 19). Then, because the triangles ABC, ABE on the same base, and between the same parallels, are equal (th. 22); and because the one triangle ABC is half the parallelogram ABCD (th. 19), the other equal triangle

*The triangles being given, with their vertices C and F taken at pleasure in the line DE, the lines BE and AD must be drawn parallel to the sides AF, BC of the triangles, to complete the parallelograms The above theorem may be proved by th. 1, and also by th. 5.

ABE is also equal to half the same parallelogram ABCD. Q. E. D.

Corol. A triangle is equal to half a parallelogram of the same base and altitude.

THEOREM XXIV.

The complements of the parallelograms which are about the diagonal of any parallelogram are equal to each other.

D G

A H

C

B

F

Let AC be a parallelogram, BD a diagonal, EIF parallel to AB and DC, and GIH parallel to AD and BC, making AI, IC complements to E the parallelograms EG, HF, which are about the diagonal DB; then will the complement AI be equal to the complement IC. For, since the diagonal DB bisects the three parallelograms AC, EG, HF (th. 19); therefore, the whole triangle DAB being equal to the whole triangle DCB, and the parts DEI, IHB respectively equal to the parts DGI, IFB, the remaining parts AI, IC must also be equal (by ax. 3). Q. E. Ď.

THEOREM XXV.

A trapezoid is equal to half a parallelogram, whose base is the sum of the two parallel sides, and its altitude the perpendicular distance between them.

A G B E

Let ABCD be the trapezoid, having its D CHF two sides AB, DC parallel; and in AB produced take BE equal to DC, so that AE may be the sum of the two parallel sides; produce DC also, and let EF, GC, BH be all three parallel to AD. Then is AF a parallelogram of the same altitude with the trapezoid ABCD, having its base AE equal to the sum of the parallel sides of the trapezoid; and it is to be proved that the trapezoid ABCD is equal to half the parallelogram AF

Now, since triangles, or parallelograms, of equal bases and altitudes, are equal (corol. 2, th. 22), the parallelogram DG is equal to the parallelogram HE, and the triangle CGB equal to the triangle CHB; consequently, the line BC bisects or equally divides the parallelogram AF, and ABCD is the half of it. Q. E. D.

THEOREM XXVI.

In any right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the other two sides.

Let ABC be a right-angled triangle, having the right angle A; then will the square of the hypothenuse BC be equal to the sum of the squares of the other two sides AC, AB. Or BC2 = AC2 + AB'.

For, on BC describe the square BE, and on AC, AB, the squares CH, BG; then draw AL parallel to BD, and join CF, AD.

F

B

D

I

K

LE

Now, because the line AB meets the two AG, AC, so as to make the sum of the two adjacent angles equal to two right angles, these two form one straight line GC (corol. 1, th. 6). And because the angle FBA is equal to the angle DBC, being each a right angle, or the angle of a square; to each of these equals add the common angle ABC, so will the whole angle or sum FBC be equal to the whole angle or sum ABD. But the line FB is equal to the line BA, being sides of the same square; and the line BD to the line BC, for the same reason; so that the two sides FB, BC, and the included angle FBC, are equal to the two sides AB, BD, and the included angle ABD, each to each; therefore the triangle FBC is equal to the triangle ABD (th. 1).

But the square BG is double the triangle FBC on