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a. 16. 1.
PROP. XXVII. THEOR. -
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 AEF, EFD equal to one another ; AB is parallel to CD.
For if it be not parallel, AB and CD being produced shall meet ei-
G demonstrated that they do not
С F meet towards AC. but those
D straight lines which meet nei5. 35. Def. ther way tho' produced ever so far are parallel b to one another. AB therefore is parallel to CD, wherefore if a straight line, &c. Q. E. D,
PROP, XXVIII. THEO K.
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 upon the two straight lines
angle EGB equal to the angle AGH, the angle AGH is equal to Book I. the angle GHD; and they are the alternate angles; therefore AB is m pvallel to CD. again, because the angles BGH, GHD are equal a. 15, 1. two right angles, and that AGH, BGH are also equal to two b. 27. 1.
c. right angles; the angles AGH, BGH are equal to the angles BGH, By Hyp.
d. 13.1. GID. take away the common angle BGH, therefore the remaining an le AGH is equal to the remaining angle GHD; and they are al. ternate angles; therefore AB is parallel to CD, wherefore if a straight line &c. Q. E. D.
PRO P. XXIX. THEOR. Fa straight line falls upon two parallel straight lines, See the it makes the alternate angles equal to one another;
notes on this
Propolition, and the exterior angle equal to the interior and opposite upon the same fide; and likewise the two interior angles upon the fame fide 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 fame side, GHD; and the two interior angles BGH, GHD upon
E the same side are together equal to two right angles.
B For if AGH belnot equal to GHD, one of them must be
greater than the other ; let AGH be the greater. and be- C H D cause the angle AGH is greater than the
F angle GHD, add to each of them the angle BGH; therefore the angles AGH, EGH are greater than the angles BGH, GHD. but the angles AGH, BGH are equal to two right a. 13. 1. 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 * together if continually produced; therefore the 12. Ax. straight lines AB, CD if produced far enough shall meet. but they see the never meet, since they are parallel by the Hypothesis. therefore the notes on this angle AGH is not unequal to the angle GHD, that is, it is equal to Proposition. it. but the angle AGH is equal to the angle EGB; therefore like-b.15.1. wise EGB is equal to GHD. add to each of these the angle BGH,
Book 'I therefore the angles EGB, BGH are equal to the angles BGH, GHD; mbut EGB, BGH are equal to two right angles ; therefore also C. 13. 1. BGH, GHD are equal to two right angles. wherefore if a straight
line &c. Q. E. D.
PRO P. XXX. THEOR.
line, are parallel to one another.
Let the straight line GHK cut AB, EF, CD; and because GHK
cuts the parallel straight lines AB, 1. 29.1. EF, the angle AGH is equal" to the angle GHF. again, because the
G straight line GK cuts the parallel A
-B straight lines EF, CD, the angle
-F and it was shewn that the angle
CAGK is equal to the angle GHF; therefore also AGK is equal to
GKD. and they are alternate an1.27. 1. gles; therefore AB is parallel b to CD. wherefore straight lines
b &c, Q. E. D.
To draw a straight line thro' 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 thro'
In BC take any point D, and join
C 2. 23. 1. line AD make 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 b. 27. 1. another, EF is parallel to BC. therefore the straight line EAF is
drawn thro' the given point A parallel to the given straight line BC. Book I. Which was to be done.
E b. 19.1.
PROP. XXXII. THEOR.
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 fides 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.
Thro' the point C draw CE parallel to the straight line AB. and a. 31.1. because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are
А equal 6. again because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior
C D 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 equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB. but the angles ACD, ACB are cqual to two right angles; therefore also the angles CBA, BAC, ACB are equal to two righe angles. wherefore if a side of a triangle &c. Q. E. D.
CoR. I. All the interior angles of any rectilineal figure, together
D with four right angles, are equal
E to twice as many right angles as
C the figure has fides. For any rectilineal figure ABCDE
F can be divided into as many triangles as the figure has sides, by drawing straight lines from a point A F within the figure to each of its
c. 13. I.
Book I. angles. And, by the preceeding Proposition, all the angles of the mtriangles are equal to twice as many right angles as there are tri
angles, that is, as there are sides of the figure, and the same anges
are equal to the angles of the figure, together with the angles at the 2. 2. Cor. point F which is the common Vertex of the triangles; that is “, to
gether with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
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 6.15.1. ABD is equal b to two right an
gles; therefore all the interior
PROP. XXXIII. T H E O R.
equal and parallel straight lines, towards the fame parts, are also themselves equal and parallel.
Let AB, CD be equal and parallel straight lines, and joined towards the fame parts by the straight A
Join BC, and because AB is pa-
D a. 19. s. are equal“; and because AB is equal to CD, and BC common to
the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other anglesb,