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AN

equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF: but let the base CB be greater than the base EF, the angle BAC 'is likewise greater than the angle EDF.

For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (4. 1.) A

D to EF; but it is not; therefore the angle BAC is not equal to the arigle EDF; neither is it less; because then the base BC would be less (24. 1.) than the base EF; but it is not; therefore the angle BAC is not less than the angle EDF; and it was shown that it is B

F not equal to it: therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D.

PROP. XXVI. THEOR. If two triangles have two angles of the one equal to two angles of the other, each to each; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to the equal angles in each; then shall the other sides be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two А

D triangles which have the angles ABC, BCA equal to the angles DEF, EFD, G viz. ABC to DEF, and BCA to EFD; also one side equal to one side; and first let those sides be equal which are adjacent to the angles that are equal in the two tri.

B

CE angles, viz. BC to EF; the other sides shall be equal, each to each, viz. AB to DE, and AC to DF; and the third angle BAC to the third angle EDF.

For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because. BG is equal

to DE, and BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal (4. 1.) to the base DF, and the triangle GBC to the triangle

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DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle GCB is equal to the angle DFE but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF, therefore the base AC is equal (4. 1.) to the base DF, and the angle BAC to the angle EDF. Next, let the sides A

D
which are opposite to
equal angles in each tri-
angle be equal to one an-
other, viz. AB to DE;
likewise in tbis case, the
other sides shall be e-
qual, AC to DF, and BC
to EF; and also the
third ongle BAC to the
Third EDF.

B
нс

E For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore (4. 1.) thé 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. Therefore, if two triangles, &c. Q. E. D.

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 A

F

B and CD being produced, do not meet towards B, D. In like manner it may be demonstrated that they

с

D 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. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXVIII. 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

A

B BGH, GHD together equal to two right angles; AB is parallel to CD. Because the angle EGB is

H н equal to the angle GHD, and C

D also (15. 1.) to the angle AGH the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel (27. 1.) to CD. Again, because the angles BGH, GHD are equal (By 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; there. føre AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.

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.); hut AB is also parallel to CD;

L therefore two straight lines

А

G are drawn through the same

B point G, parallel to CD, and

K yet not coinciding with one another, which is impossible c (11, Ax.) The angles AGH

-D

H GHD therefore are not une equal, that is, they are equal to one another. Now, the

F angle EGB is equal to AGH (15. 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 (13. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles. "Wherefore, if a straight line, &c. Q. E. D.

Cor. 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 a 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.

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 angle AGH is equal

are

(29. 1.) to the angle GHF: Again, because the straight line

A GK cuts the parallel straight

B lines EF, CD, the angle GHF is equal (29. 1.) to the angle

E GKD; and it was shown that

H the angle AGK is equal to the angle GHF; therefore also AGK

K is equal to GKD; and they are

C alternate angles; therefore AB

D js parallel (27.1.) to CD. Wherefore straight lines, &c. Q. E. D.

PROP. XXXI. PROB. 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

A through the point A, parallel to

F the straight line BC.

In BC take any point D, and join AD; and at the point A,

B D

с in the straight line AD, make (23. 1.) the angle DAE equal to the angle ADC; 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 atraight line EAF is drawn through the given point A parallel to the given straight line BC: Which was to be done.

PROP. XXXII. THEOR. If a side of any triangle be produced, the exterior an gle 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

А CE parallel (31. 1.) 10 the straight line AB; and because

E AB is parallel to CE and AC meets thein, the alternate an: gles BAC, ACE are equal (29.

B 1.) Again; because AB is

D

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