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AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base cb is 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 (1. 4.) to EF; but it is not, therefore the angle Bac is not equal to the angle EDF; neither is it less; because then the base BC would be less (1. 24.) than the base EF; but it is not; therefore

E the angle BAC is not less than B the angle EDF; and it was shown that it is 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 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 equal angles in each ; then shail the other sides be equal each each : and also the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, 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 triangles, viz. BC

A

D 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 a B be not equal to DE, one of them must be the greater. Let

F AB be the greater of the two, and make BG equal to DE, and join gc; therefore, because BG is equal to DE, and Bo to EF (Hyp.), 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 (1. 4.) to the base DF, and the triangle GBC to the triangle DEF,

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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 asо 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; the base therefore AC is equal (1. 4.) to the base DF, and the third angle BAC to the third angle EDF. Next, let the sides which are A

D opposite to equal angles in each triangle be equal to one another, viz. AB TO DE; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the third angle BAC to B HC E the third EDF.'

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 Bu is equal to EF, and AB to DE (Hyp.), the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles (Hyp.); therefore the base au is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be 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 (Hyp.); therefore also the angle bha is equal to the angle BCA, that is, the exterior angle bha of the triangle Auc is equal to its interior and opposite angle BCA, which is impossible (1. 16.); wherefore bc is not unequal to EF, that is, it is equal to it; and AB is equal to DE (Hyp.); therefore the two A B, 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 shall be parallel.

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

For, if it be not parallel, AB and cp 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 (1. 16.) than the interior and opposite angle EFG; but it is also equal A

B toit (Hyp.), which is impossible:

G therefore A B and cd being pro- C. duced do not meet towards B, D. In like mannerit may be demonstrated that they do not meet towards AC; but those straight lines which meet neither way, though produced ever so far, are parallel (Def. 35.) to one another. AB therefore is parallel to CD. Wherefore if a straight line, &c. Q. E. D.

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PROP. XXVIII. THEOR.

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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 shall be parallel to one another. Let the straight line EF, which falls

E upon the two straight lines A B, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon

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A same side; or make the interior angles on the same side BGH, GHD together

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-D equal to two right angles; A B is parallel to CD.

F Because the angle EGB is equal to the angle GhD, and the angle EGB equal (1. 15.) to the angle AGH, the angle Agh is equal to the angle Ghd, and they are the alternate angles ; therefore AB is parallel (1. 27.) to CD. Again, because the angles BGH, GAD are equal (by Hyp.) to two right angles, and that AGH, BGH are also equal (1. 13.) 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 (1. 27.). Wherefore if a straight line, &c. Q. E. D.

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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 egual to two right angles.

Let the straight line EF fall upon the parallel straight lines A B, 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, GHV upon the same side are together equal to two right angles.

For, if Agh be not equal to GHD, one of them must be greater than the other; C let Agh be the greater; and because the angle agh is greater than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles AGH, Bgh are equal (1. 13.) to two right angles ; therefore the angles BGH, GHD are less than two right angles; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, do meet (Ax. 12.) together if continually produced; therefore the straight lines AB, CD, if produced far enough, shall meet; but they never meet, since they are parallel by the hypothesis ; therefore the angle Agh is not unequal to the angle GHD, that is, it is equal to it; but the angle Agh is equal (1. 15.) to the angle EGB; therefore likewise EGB is 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 (1. 13.) to two right angles; therefore also BGH, GhD are equal to two right angles. Wherefore, if a straight line, &c. Q.E.D.

PROP. XXX. THEOR. Straight lines which are parallel to the same straight line 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 gak cut AB, EF, CD; and because GHK cuts the parallel straight lines A B, EF, the angle agh is equal (I. 29.) A to the angle GHF. Again, because the

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F straight line gak cuts the parallel straight lines EF, CD, the angle GhF K

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D is equal (1. 29.) to the angle GKD; and it was shown that the angle AGK is equal to the angle GhF; therefore also AGK is equal to GKD; and they are alternate angles; therefore A B is parallel (1. 27.) 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 the E

F straight line BC.

In Bc take any point D, and join BAD; and at the point a, in the straight line AD, make (I. 23.) 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 (1. 27.) to Bc. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC. Which was to be done.

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PROP. XXXII. THEOR.

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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 triangles, viz. ABC, BCA, CAB, are together equal to two right angles.

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