Join BC; and because A B and C D are parallel the alternate angles A B C, BCD are equal; and because A B is equal to CD, and BC common to the two triangles ABC, BC D, and the angle A B C is equal to the angle B CD; therefore the base A C is equal to the base B D, and the triangle A B C to the triangle B C D, and the angle ACB opposite A B to the angle CBD opposite CD. And because the straight line DC meets the two A C, BD, making the alternate angles AC B, CBD equal to one another, A C is parallel to B D.

XXXVI.-EUCLID I. 34.

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, or divides them into two equal parts.

Let ABCD (Fig. 28) be a parallelogram of which B C is a diameter; the opposite sides and angles of the figure are equal to each other, and the diameter B C bisects it.

Because A B is parallel to CD, and BC meets them, the alternate angles A B C, B C D are equal; and because A C is parallel to B D, and B C meets them, the alternate angles ACD, CBD are also equal; wherefore the two triangles A CB, CBD have two angles ABC, BCA in one, equal to two angles B CD, CBD in the other, each to each; and one side B C common to the two triangles, which is adjacent to their equal angles. There

fore the triangle B A C is equal to the triangle BDC, and the sides A B, A C to the sides CD, DB respectively, and the angle CAB to the angle B D C. Also, because the angle ABC is equal to BCD, and CBD to AC B, the whole angle ABD is equal to the whole angle AC D. Thus, all the opposite sides and angles are equal respectively, and it was shewn that the triangle ABC is equal to the triangle B CD; therefore, the diameter BC divides the parallelogram into two equal parts.

XXXVII.

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

Let the parallelograms A B C D, E B C F (Fig. 29) be upon the same base B C, and between the same parallels AF, BC; they shall be equal to each other.

Because ABCD is a parallelogram, AB is parallel and equal to DC; and because EBC F is a parallelogram, E B is equal and parallel to CF; therefore the two, AB, BE, are equal to the two DC, CF, respectively, and the angle ABE is equal to the angle D C F (Prop. 10), wherefore the triangle A B E is equal to the triangle DCF. Take the triangle A B E from the rectilineal figure A B CF, and from the same figure take the equal triangle DCF; therefore the remainders are

equal, that is the parallelogram A B C D is equal to the parallelogram EBC F.

XXXVIII.-EUCLID I. 36,

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

Let A B C D, E F G H (Fig. 30) be parallelograms upon equal bases, BC, FG, and between the same parallels, A H, B G. They are equal to each other.

Join B E, CH. And because B C is equal to F G and F G to EH, B C is equal to E H; but B C and E H are parallels joined towards the same parts by the straight lines BE, CH. Therefore (Prop. 35) B E, C H are themselves equal and parallel, and E B C H is a parallelogram, and it is equal to A B C D because it is on the same base B C, and between the same parallels. For a like reason it is equal to the parallelogram E H G F; wherefore E H G F is also equal to A B C D.

XXXIX.-EUCLID I. 37.

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

Let the triangles A B C, D BC (Fig. 31) be upon the same base B C, and between the same parallels AD, BC; the triangle A B C is equal to D B C.

Produce A D both ways to E, F, and through B let B E be parallel to CA; and in like manner

E

let C F be parallel to CD. Therefore each of the figures E B C A, D BCF is a parallelogram; and EBCA is equal to D B C F, because they are upon the same base B C, and between the same parallels B C, E F; and the triangle A B C is half of the parallelogram E B C A, and the triangle DBC is half of the parallelogram D B C F (Prop. 37); wherefore the triangle A B C is equal to the triangle D B C.

.

XL.

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

Let the triangles A B C, D E F (Fig. 32) be upon equal bases B C, E F, and between the same parallels B F, A D. The triangle A B C is equal to DE F.

Produce A D both ways to G and H, and let B G be parallel to A C, F H parallel to ED: then each of the figures G B CA, DEFH is a parallelogram, and they are equal to each other because they are upon equal bases B C, E F, and between the same parallels B F, G H; and the parallelogram GABC is double of the triangle ABC, and D E F H is double of DEF; wherefore the triangle A B C is equal to D E F.

XLI.

If two triangles have one side, and the adjacent

angle of the one equal to one side and the adjacent angle of the other, and if the triangles are equal the other sides are also equal.

Let ABC, DEF (Fig. 32) be two triangles, having the angle B A C equal to the angle E D F, and the adjacent side A B equal to D E, and let the triangle A B C be equal to the triangle ED F. Then the other sides B C, AC are respectively equal to E F, DF.

Let the triangle ABC be super-imposed on DEF so that A B shall coincide with D E. Then, because the angle B A C is equal to ED F, the line A C will lie upon E F, and let the point C coincide with the point G in E F (produced if necessary). Join E G. Then the triangle D EG will be equal to the sum or difference of the triangles D E F, E F G, according as G falls without or within the base D F; that is, according as D G or AC is greater or less than D F; and, therefore, if A C is greater than D F, the triangle DEG or A B C is greater than D E F, and if AC is less than DF, the triangle A B C is less than D E F. Hence, conversely, if triangle A B C is not greater than D E F, A C is not greater than DF; and if A B C is not less than DEF, AC is not less than D F; and, therefore, if the triangle A B C is neither greater nor less than D E F, A C is neither greater nor less than D F; that is to