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PROPOSITION XXXI. THEOR.

If one side of a triangle be produced, the exterior angle is equal to the two interior opposite angles.

Let BC, one of the sides of the triangle ABC, be produced to D, the exterior angle ACD is equal to the interior opposite angles ABC and BAC.

A

For the three interior angles A, B, C, are together equal to two right angles (30), but the two angles ACB and ACD are also (5) together equal to two right angles; therefore the two angles ACB, ACD are equal to three A, B, C; take away the common angle C, or ACB, and there remains the angle ACD, equal to the angles A and B, or the exterior angle, equal to the two interior opposite angles.

B

-D

C

PROPOSITION XXXII. THEOR.

If two triangles be equiangular, and if a side of the one.be equal to a side of the other which is opposite to an equal angle, the triangles are equal.

Let ABC and DEF be equiangular triangles, and let the side AB opposite to the angle C in the one be equal to DE opposite to F, an equal angle in the other, the triangle ABC is in all respects equal to the triangle DEF.

For if BC be not equal to EF, A

let BC be greater, and make (2) BG equal to EF, and join AG. Because the side BG is equal to EF, and AB to DE, and the included angles B and E are B equal, therefore the triangles ABG and DEF are equal, and the angle BGA is equal to EFD, but EFD is equal to BCA;

GCHE

F

therefore the angle BGA is equal to BCA, the exterior angle of the triangle ACG equal to its interior opposite angle, which is (16) impossible; therefore BC is not greater than EF. If BC be produced to H, and AH joined, it may be demonstrated in the same manner that BC is not less than EF; and it has been already proved that it is not greater; therefore BC is equal to EF. Since therefore the side AB is equal to DE, and BC to EF, and the included angles B and E are equal, the triangle ABC is equal (12) to DEF, their areas are equal, and the side AC is equal to DF.

PROPOSITION XXXIII. PROB.

Through a given point to draw a straight line 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 line BC.

E

B D

A

F

In BC take any point D, and join AD, and at the point A in the straight line AD make (27) the angle DAE equal to the angle ADC, and produce the straight line EA to F, EF is parallel to AB. Because the straight line. AD which meets the two straight lines BC and EF makes the alternate angles EAD and ADC equal; EF is (Def. 12) parallel to BC; therefore the straight line EAF is drawn through the given point, A parallel to BC.

PROPOSITION XXXIV. THEOR.

If a straight line, intersecting two other straight lines makes the exterior angle equal to the interior opposite angle on the same side of the line; or makes the interior angles on the same

side together equal to two right angles, the two straight lines are parallel.

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

G

H

F

B

D

Because the angle EGB is equal to the angle GHD, and also (7) to the angle AGH, the angle AGH is equal to GHD, and they are alternate angles; therefore AB is parallel to CD. Again, because the angles BGH and GHD are equal to two right angles, and AGH, BGH are also equal (5) to two right angles, the angles AGH and BGH are equal to the angles BGH, GHD; take away the common angle BGH, and the remaining angle AGH is equal to the remaining angle GHD, and they are alternate angles; therefore (Def. 12) AB is parallel to CD.

Cor. Straight lines which are perpendicular to the same straight line are parallel.

PROPOSITION XXXV. THEOR.

If a straight line intersect two parallel straight lines it makes the alternate angles equal, the exterior angle equal to the interior opposite angle on the same side of the line, and the two interior angles on the same side together equal to two right angles.

Let AB and CD be two parallel straight lines, and let any straight line EF intersect them, the alternate angles AGH and GHD are equal, the exterior angle EGB is equal to GHD, the interior opposite angle on the same side of EF, and the two interior angles BGH, GHD on the same side, are together equal to two right angles.

K G

E

M

H

L

K

B

M

H

D

F

It is evident that the line KL, with which (Def. 12) the lines AB, CD make A equal alternate angles, may be so drawn as to intersect EF, either between the lines AB, CD, or beyond them. First, let KL intersect EF between the lines in the point M. Because AB and CD A make equal alternate angles with KL, the angle GKM is equal to HLM, and the angle GMK is equal to LMH (7), the third angles of the triangles KGM and MHL are (30, Cor. 1) equal; hence the angle KGM is equal to MHL, or the angle AGH is equal to GHD, and therefore EF also makes equal alternate angles with AB and CD. Next, let the line KL cut the line EF beyond AB in the point K. Because the lines AB, CD make equal alternate angles with KL, the angle AML is equal to MLH, but the angle AML is also equal to KMG; therefore KMG is equal to MLH; to each of these add the angle K, common to the two triangles KGM and KLH, and the two angles KMG and MKG are equal to the two KLH and LKH, but KMG and MKG are equal (31) to the angle MGH and KLH, and LKH to GHD; therefore the angle MGH or AGH is equal to GHB, and these are alternate angles; wherefore the lines AB, CD make equal alternate angles with EF. But the angle AGH is equal (7) EGB; therefore the angle EGB is equal to GHD, the exterior to the interior opposite angle; add to each of these the angle BGH, and the angles EGB, BGH are equal to BGH and GHD, but EGB and BGH are (5) equal to two right angles; therefore BGH and GHD are also equal to two right angles.

Cor. A straight line perpendicular to one of two parallels is also perpendicular to the other.

PROPOSITION XXXVI. THEOR.

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

Let AB and CD be each of them parallel to EF, then AB is also parallel to CD.

A

C

K

B

H

D

Let the straight line GHK cut the lines AB, EF, CD, and because GHK cuts the parallel straight lines AB and E EF, the angle AGH is equal (35) to GHF. Again, because GK cuts the straight lines EF and CD, the angle GHF is equal to (35) the angle GKD; and it was proved that the angle AGK is equal to GHF; therefore AGK is equal GKD, and they are alternate angles; therefore AB is parallel (Def. 12) to CD.

PROPOSITION XXXVII. THEOR.

Straight lines parallel to the sides of an angle contain an equal angle.

E

Let the straight lines AB, AC be parallel to DE, DF, the sides of the angle EDF, the angle BAC is equal to EDF, whether the angles lie within or wholly without one another. Through the vertices of the angles draw the straight line GAD, since AC is parallel to DF, the exterior angle GAC is equal (35) GDF, the interior opposite angle; and, for the same reason, the angle GAB is equal

G

B

A

C

с

E

F

to GDE; and by adding equals to equals, or taking equals from equals, the whole or the part BAC is equal to the whole or the part DEF.

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