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for the right angles B and E are equal, the side AB=DE,

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D and BH=EF; hence, these triangles are equal (Th. 1.), and consequently AH=DF. Now (by hyp.), we have DF=AC; and therefore, AH=AC. But by the last proposition, the oblique line AC cannot be equal to the oblique line AH, which lies nearer to the perpendicular AB; B

нс E therefore it is impossible that BC can differ from EF: hence, then, the triangles ABC and DEF are equal.

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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

E

B towards A, C; let them be produced and meet towards B, D in the point G; therefore GEF

G is a triangle, and its exterior

D angle AEF is greater (9. 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 (15 Def.) to one another. AB therefore is parallel to CD.

PROP. XX. THEOR.

If a straight line falling upon two other straight lines makes the exterior

angle 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.

Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to GHD, the interior and opposite angle upon the same side ; or let it make the interior angles on the same side BGH, GHD together equal to two right angles ; AB is parallel to CD.

Because the angle EGB is equal to the angle GHD, and also (8. 1.) to

E the angle AGH, the angle AGH is equal to the angle GHP; and they A

B are the alternate angles; therefore AB is parallel (19. 1.) to CD. Again, because the angles BGH, GHD are equal (by Hyp.) to two right angles,

H

D and AGH, BGH, are also equal (6. 1.) to two right angles, the 'angles AGH,

F 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. XXI. THEOR.

GHD;

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,

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 Α.

G

-В will be parallel to CD (19. 1.);

K but AB is also parallel to CD; therefore two straight lines are drawn through the same point

C

H G, parallel to CD, and yet not coinciding with one another,

A 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 (8. 1.); and AGH is proved to be equal to GHD; therefore EGB is likewise 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 (6. 1.) to two right angles ; therefore also BGH, GHD are equal to two right angles.

Cor. 1. If BGH is a right angle, GHD will be a right angle also;

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therefore every line perpendicular to one of two parallels, is perpendicular to the other.

Cor. 2. 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 ob tuse, 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 have already named interior angles on the same side ; AGH, GỨC, 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.

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PROP. XXII. THEOR.

If a straight line, cutting two other lines, make the sum of the two interior

angles on the same side, less than two right angles, those two lines will meet if produced.

Let the straight line EF (see last figure,) make with the two lines KL and CD, 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 (6. 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. Through a given point G, no more than one line can be drawn parallel to a given line CD. For there is but one line AB, which makes the sum of the two angles AGH, GHC, equal to two right angles. Every other line, KL for example, would make the sum of the two interior angles less or greater than two right angles ; therefore it would meet CD.

PROP. XXIII. 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.

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Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (21. 1.) to the angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal (21. 1.) to the angle GKD; and it was shewn that the angle AGK is equal to the angle GHF; therefore also AGK is equal to GKD; and they are alternate angles ; therefore AB is parallel (19. 1.) to CD.

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PROP. XXIV. THEOR.

Two angles are equal if their sides be parallel, each to each, and lying in the

same direction.

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If the straight lines AB, AC be parallel to DF, DE; the angle BAC is equal to EDF.

For, draw GAD through the vertices. I And since AB is parallel to DF, terior angle GAB is (21. 1.) equal to

H GDF;

nd, for the same reason, GAC is equal to GDE ; there consequently remains the angle BAC=EDF.

D
Cor. If BA, AC be produced to I and
H, the angle BAC=HAI ; hence, the
angle HAI is also equal to EDF.

SCHOLIUM. The restriction of this proposition to the case where the side AB lies in the same direction with DF, and AC in the same direction with DE, is necessary ; because the angle CAI would have its sides parallel to those of the angle EDF, but would not be equal to it. In that case, CAI and EDF would be together equal to two right angles.

PROP. XXV. 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. ABC, BCA, CAB, are together equal to two right angles.

Through the point C draw CE

A parallel to the straight line AB; and because AB is parallel to

E CE, and AC meets them, the alternate angles BAC, ACE are equal (21. 1.) Again, because AB is parallel to CE, and BD B

С

D 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 CBĂ, BAC, ACB; but the angles ACD, ACB are equal (6. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles.

Cor. 1. 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. 2. 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. 3. 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 right angle.

Cor. 4. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.

Cor. 5. Since every equilateral triangle (Cor. 3. 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.

PROP. XXVI. THEOR.

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 to

D 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

E triangles ; that is, (2 Cor. 8. 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, together with four right angles, that is, the an A

B

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