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hence BGH+GHD is equal to two right-angles; therefore the lines AB, CD are parallel. (Prop. 20.) In like manner, if the alternate angles AGH, GHD are equal, the lines AB, CD will be parallel.

Again, by hypothesis, the angle EGB=GHD; and EGB= AGH; (Prop. 4;) hence AGH=GHD, and they are alternate angles; therefore AB, CD are parallel. In like manner, if the exterior angle EGA GHC the interior opposite angle, the lines AB, CD will be parallel. Hence,

Any two straight lines are parallel, which, meeting a third line, make the alternate angles equal; or which make the opposite exterior and interior angles equal.

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If two straight lines AI, BD, meeting a third AB, make the sum of the two interior angles BAI, ABD, less than two rightangles, the lines AI, BD, will meet if produced.

Draw the straight line AC, making the angle CAB=ABF, in other words, so that the angles CAB, ABD, taken together, may be equal to two right-angles; and complete the rest of the construction, as in Prop. 20th. Since AEK is a rightangle, AE the perpendicular is shorter than AK the oblique line; hence, (Prop. 13,) in the triangle AEK, the angle AKE, opposite the side AE, is

E

C

I

D

K

F

LN

H

less than the right-angle AEK, opposite the side AK. But AKE=IKF; (Prop. 4;) therefore IKF is less than a rightangle; hence the lines KI and FD will meet if produced. (Prop. 19.) Hence,

Any two straight lines, which, meeting a third line, make the sum of the two interior angles on the same side less than two right-angles, will meet if produced.

Cor. Through a given point A, only one line can be drawn parallel to a given line BD. For there is but one line AC, which makes the sum of the two angles BAC, ABD equal to two right-angles; and this is the parallel required. Every other line AI, &c., would make the sum of the interior angles greater or less than two right-angles; therefore it would meet BD.

Scholium. The two lines IL and DH, will meet on the side of AB, on which the sum of the two interior angles is less than two right-angles.

PROPOSITION XXIII. THEOREM.

If two parallels AB, CD, are intersected by a third line EF, the sum of the interior angles AGH, GHC, will be equal to two right-angles.

For, if the sum of angles AGH+ GHC, were greater or less than two right-angles, the two straight lines AB, CD, would meet on one side of EF or the other; (Prop. 22 ;) consequently they would not be parallel,

A

E

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B

C

D

H

F

which is contrary to the hypothesis. Hence,

If two parallels are intersected by a third line, the sum of the two interior angles on the same side, will be equal to two right-angles.

Cor. If GHC is a right-angle, AGH will be a right-angle also; therefore, every line perpendicular to one of two parallels will be perpendicular to the other.

PROPOSITION XXIV. THEOREM.

If two parallels AB, CD, are intersected by a third line EF, the alternate angles BGH, GHC, will be equal; also the opposite exterior and interior angles EGB, GHD, will be equal.

For, since the sum of the angles GHC, GHD is equal to two rightangles, (Prop. 1,) and BGH, GHD A is also equal to two right-angles, (Prop. 23,) we have GHC+GHD C =BGH+GHD. Taking GHD

E

B

D

H

F

from each, there will remain the alternate angle GHC= BGH. In like manner it may be proved that the alternate angles AGH, GHD, are equal.

Again, the angle EGB=AGH, (Prop. 4,) and GHD= AGH as proved above, hence EGB=GHD. (Ax. 1.) In like manner it may be proved that EGA=GHC. Hence,

If any two parallels are intersected by a third line, the alternate angles will be equal; and the opposite exterior and interior angles will also be equal.

Cor. The sum of all the angles formed by the intersection of two parallels by a third line, is equal to eight right-angles. If the third line cuts the parallels obliquely, the four acute angles will be equal to each other, also the four obtuse will be equal to each other; and if one of the acute angles be added to one of the obtuse, the sum will always be two right-angles.

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Two lines AB, CD, parallel to a third, are parallel to each other.

Draw PQR, perpendicular to EF, and cutting AB, CD. Since AB is parallel to EF, PR will be perpendicular to AB; (Prop. 23. Cor.;) and since CD is parallel to EF, PR will Afor a like reason be perpendicular to

CD. Hence AB and CD are per

E

R

C

-D

Q

B

pendicular to the same straight line; therefore, (Prop. 18,) they are parallel. Hence,

Any two straight lines which are parallel to a third line, are, parallel to each other.

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Two parallels AB, CD, are every where equally distant.

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lines will at the same time be perpendicular to CD. (Prop. 23. Cor.) And we are now to show that they will be equal to each other.

Now, if GF be joined, the angles GFE, FGH, will be alternate angles, and therefore, (Prop. 24,) equal to each other. Also, because the straight lines EG, FH are perpen

dicular to the same straight line AB, and consequently parallel, the angles EGF, GFH, will be alternate interior angles, and therefore equal. Hence the two triangles EFG, FGH have a common side, and two adjacent angles in each equal; hence these triangles are equal; (Prop. 6;) therefore the side EG, which measures the distance of the parallels AB and CD at the point E, is equal to the side FH, which measures the distance of the same parallels at the point F. Hence, Any two parallels are every where equally distant.

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If two angles BAC, DEF have their sides parallel, each to each, and lying in the same direction, those angles will be equal.

Produce ED, if necessary, till

B

D

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it meets AC in G. Now since
EF and GC are parallel and
ED intersects them, the angle
DEF=DGC; (Prop. 24;) and
since the line DG is parallel to H
AB, the angle DGC-BAC:
hence the angle DEF BAC. (Ax. 1.)

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If any two angles have their sides parallel, and lying in the same direction, these two angles are equal.

Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, is necessary, because if FE were produced toward H, the angle DEH would have its side parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC, would be together equal to two right-angles. For BAC=AGE; (Prop. 24;) adding DEH to each, BAC+

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