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Straight lines which join the adjacent extremities of two equal and parallel straight lines are themselves equal and parallel.

Let AB and CD be equal and parallel straight lines, and let their adjacent extremities A, C, and B, D be joined, AB and CD are also equal and parallel.

Join BC, and because AB is parallel to DC, A and BC meets them, the alternate angles ABC, BCD are (35) equal; therefore in the two triangles ABC and DCB the side AC is equal

C

B

D

to DC, and BC common to both, and the included angle ABC is equal BCD; therefore the triangles (12) are equal, and AC is equal to BD, but the angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC and BD, and makes the alternate angles ACB, CBD equal, AC is (Def. 12) parallel to BD, and it was proved to be equal to it; wherefore AC is equal and parallel to BD.

Cor. Parallel lines are equidistant. For if ACD and BDC were right angles, AC and BD would be equal perpendiculars and hence the lines are equidistant.

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If two straight lines make with a third straight line the two interior angles on one side of it together less than two right angles, these two lines shall meet on that side of the third on which the two angles are that are together less than two right angles.

Let the straight lines AB, CD make with EF the two interior angles AEF and EFC on one side of EF together less than two right angles, AB, CD shall meet towards A and C.

A

B

H

D

Through the point E draw (33) the straight line GH parallel to CD. Then, if AB, CD G do not meet towards A and C, they must either be parallel and not meet, or they must meet on the other side of EF. But they are not parallel, because the angles AEF and EFC would then be equal to two right angles, which they are not; neither do they meet on the other side of EF, for the angles BEF, EFD would then be two angles of triangle, and less than two right angles, but this is impossible; for the four angles AEF, EFC, BEF, EFD are together equal (5) to four right angles, of which the two AEF and EFC are (Hyp.) less than two right angles; therefore the other two BEF, EFD are greater than two right angles; therefore since AB, CD are not parallel, and since they do not meet towards B and D, they must meet if produced towards A and C.

Cor. Two straight lines which intersect one another cannot be both parallel to the same straight line.

PROPOSITION XL. THEOR.

The opposite sides and angles of a parallelogram are equal, and the diagonal bisects it.

Let ABCD be a parallelogram, the opposite sides and angles of the figure are equal to one another, and the diagonal CD bisects it, that is, divides it into two equal parts.

B

D

Because AB is parallel to CD, and BC meets A them, the alternate angles ABC and BCD are (35) equal; and because AC is parallel to BD, and BC meets them, the alternate angles ACB c and CBD are equal; wherefore the two triangles ABC and CBD have two angles ABC, BCA in the one, equal to the two angles BCD, CBD in the other; the triangles therefore (30, Cor. 1) are equiangular, and the side BC which is opposite

to an equal angle in each is common to both, the triangles are therefore (29) every way equal; the side AB is equal to the side CD, and AC to BD, and the angle BAC to the angle BDC; and because the angle ABC is equal to the angle BCD, and CBD to ACB, the whole angle ABD is equal to the whole angle ACD; wherefore the opposite sides and angles of a parallelogram are equal, and also the diagonal bisects it; for, as already proved, the triangle ABC is equal to the triangle CBD.

PROPOSITION XLI. THEOR.

The diagonals of a parallelogram bisect one another.

Let ABCD be a parallelogram, the diagonals AC and BD bisect one another.

B

D

E

Let AC and BD intersect in the point E. A Because AD is parallel to BC, and AC meets them, the angle EAD is equal to (35) the alternate angle ECB, and also the vertical angles AED and BEC are (7) equal, the triangles AED and BEC have the two angles EAD, AED in the one equal ECB, BEC in the other, and are therefore equiangular, and the sides AD and BC which are opposite to equal angles in each are equal; wherefore the triangles are every way equal; AE is equal to EC, and BE to ED.

PROPOSITION XLII. PROB.

To make a rectangle that shall have its two adjacent sides equal to two given straight lines.

Let AB and C be two given straight lines, it is required to make a rectangle that shall have its two adjacent sides equal to AB and C.

From A draw AD (22) perpendicular to AB, and make (2) AE equal to C, through E draw (33) EF parallel to AB,

and

D

F

through B draw BF parallel to AE. Since
the opposite sides are parallel, EABF is E
(Def. 26) a parallelogram; and since the
angle A is a right angle, it is (Def. 29) a
rectangle, and its two adjacent sides are A
equal to the two straight lines AB and C.

B

C

PROPOSITION XLIII. THEOR.

Every oblique parallelogram is equal (Def. 38) to a rectangle upon the same base and between the same parallels.

Let ABCD be a parallelogram, and EBCF a rectangle on the same base BC, and between the same parallels BC, ED the parallelogram is equal to the rectangle.

Because (40) the side AB is equal to

CD, and the side BE to CF, and the angles A
E and F are right angles, the two right-
angled triangles ABE and DCF have the
hypothenuse AB and the side BE in the
one equal to the hypothenuse DC, and CF

FD

a like side in the other, and therefore the triangles are every way equal; to each of these add the space ABCF, and the parallelogram ABCD is equal to the rectangle EBCF.

PROPOSITION XLIV. THEOR.

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

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF and BC, the parallelograms are equal (Def. 38).

For the parallelogram ABCD is equal (43) to a rectangle upon the same base, and between the parallels and the paral

lelogram EBCF is equal to the same rectangle, and (Ax. 1) things which are equal to the same thing are equal to one another; therefore the parallelograms ABCD and EBCF are equal.

B

DE

Cor. Triangles upon the same base and between the same parallels are equal to one another; for the triangle ABC is (40) half the parallelogram ABCD, and the triangle FCB is half the parallelogram EBCF; and therefore (Ax. 7) the triangle ABC is equal to the triangle FCB.

PROPOSITION XLV. THEOR.

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

Let ABCD, EFGH be two parallelograms on equal bases BC, FG, and between the same parallels AH, BG, the parallelogram ABCD is equal to EFGH.

C

DE

F

G

Join BE and GH, and because BC is equal to FG, and FG (40) to EH, BC is equal to EH, and they are parallels, and have their adjacent extremities joined by the lines BE, CH; therefore (39) BE, CH are both equal and paral- B lel, and EBCH is a parallelogram, and it is equal to ABCD, because it is on the same base BC (44), and between the parallels BC, AH. For the same reason, the parallelogram EFGH is also equal to EBCH; therefore the parallelogram. ABCD is likewise equal to the parallelogram EFGH.

Cor. Triangles upon equal bases and between the same parallels are equal to one another. For the diagonals AC, EG bisect the parallelograms, and therefore (Ax. 7) the triangle ABC is equal to the triangle EFG.

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