the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles are equal, each to each, to which the equal sides are opposite; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA, which is impossible (16. 1.); wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore the two, AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF.

PROP. XXVII. THEOR.

If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines are parallel.

Let the straight line EF, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C; let them be produced and meet towards B, D in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater (16. 1.) than the interior and opposite angle EFG; but it is also equal to it, which is impossible: therefore, AB and CD being produced, do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (30. Def.) to one another. AB therefore is parallel to CD.

A

E

B

G

C

F

D

If a straight line falling upon two other straight lines makes the exterior an gle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines are parallel to one another.

angle AGH, the angle AGH is equal to the angle GHD; and they are th alternate angles; therefore AB is parallel (27. 1.) to CD. Again, because the angles BGH, GHD are equal (hyp.) to two right angles, and AGH, BGH, are also equal (13. 1.) to two right angles, the angles AGH, BGH are equal to the angles BGH, GHD: Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD.

COR. Hence, when two straight lines are perpendicular to a third line, they will be parallel to each other.

PROP. XXIX. THEOR.

If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles.

For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be parallel to CD (27.1.); but AB is also paral

lel to CD; therefore two straight lines are drawn through the same point G, parallel to CD, and yet not coinciding with one another, which is impossible (11. Ax.) The angles AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH (15. 1.); and AGH is proved to be equal to GHD; therefore EGB is like

wise equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles.

COR. 1. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles.

For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet on the other side of EF; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles (13 1.) of which the two, KGH, CHG, are by supposition less than two right angles; therefore the other two, HGL, GHD are greater than two right angles. Therefore, since KL and CD are not parallel, and since they do not meet towards L and D, they must meet if produced towards K and C.

COR. 2. If BGH is a right angle, GHD will be a right angle also; therefore every line perpendicular to one of two parallels, is perpendicular to the other.

COR. 3. Since AGE=BGH, and DHF=CHG; hence the four acute angles BGH, AGE, GHC, DHF, are equal to each other. The same is the case with the four obtuse angles EGB, AGH, GHD, CHF. It may be also observed, that, in adding one of the acute angles to one of the obtuse, the sum will always be equal to two right angles.

SCHOLIUM.

The angles just spoken of, when compared with each other, assume different names. BGH, GHD, we have already named interior angles on the same side; AGH, GHC, have the same name; AGH, GHD, are called alternate interior angles, or simply alternate; so also, are BGH, GHC : and lastly, EGB, GHD, or EGA, GHC, are called, respectively, the opposite exterior and interior angles; and EGB, CHF, or AGE, DHF, the alternate exterior angles.

PROP. XXX. THEOR.

Straight lines which are parallel to the same straight line are parallel to one another.

Let AB, CD, be each of them parallel to EF; AB is also parallel to

CD.

Let the straight line GHK cut AB, EF, CD; and because GHK cuts

the parallel straight lines AB, EF, the

To draw a straight line through a given point parallel to a given straight

line.

Let A be the given point, and BC the given straight line, it is required to draw a straight line through the point A, parallel to the straight line

BC.

A

In BC take any point D, and join AD; and at the point A, in the straight line AD, make (23.1.) the angle DAE equal to the angle ADC;

E

F

and produce the straight line EA to F.

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel (27. 1.) to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

PROP. XXXII. THEOR.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles

Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. AВС, ВСА, CAB, are together equal to two right angles.

Through the point C draw CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (29. 1.) Again, because AB is parallel to CE, and BD falls upon

them, the exterior angle ECD is equal to the interior and opposite angle ABC, but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BАС, АСВ; but the angles ACD, ACB are equal (13. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles.

COR. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition,

all the angles of these triangles are equal

D

to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are. equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, (2 Cor. 15. 1.) together with four right angles. Therefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, to

gether with four right angles that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four.

COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles.

Because every interior angle ABC, with its adjacent exterior ABD, is equal (13. 1.) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles.

H

I

A

C

D

B

COR. 3. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

COR. 4. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.

COR. 5. In any triangle there can be but one right angle; for if there were two, the third angle must be nothing. Still less can a triangle have more than one obtuse angle.

COR. 6. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.

COR. 7. Since every equilateral triangle (Cor. 5. 1.) is also equiangular, each of its angles will be equal to the third part of two right angles; so that if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by of one right angle.

COR. 8. The sum of the angles in a quadrilateral is equal to two righ angles multiplied by 4-2, which amounts to four right angles; hence, if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanctions the Definitions 25 and 26, where the four angles of a quadrilateral are said to be right, in the case of the rectangle and the square.

COR. 9. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles; hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or of one right angle.

COR. 10. The sum of the angles of a hexagon is equal to 2x (6-2), or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle.

SCHOLIUM.

When (Cor. 1.) is applied to polygons, which have re-entrant angles, as ABC each re-entrant angle must be regarded as greater than two right angles.