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PROPOSITION XXVIII. THEOREM.
If a straight line falling on two other straight lines make the exterior angle equal to the interior and opposite on the same side of the line; or make the interior angles on the same side together equal to two right angles; the two straight lines shall be parallel to one another.
Let the st. line EF, which falls on the two st. lines AB, CD, make the ext. _ EGB = int. and opp. LGHD on the same side of EF; or make the two int. Ls BGH, GHD on the same side together = 2 rt. Zs.
Then AB shall be || CD.
Dem. : LEGB=LGHD,(hyp.)and LEGB=LAGH,
(i. 15) is LAGH= LGHD; (ax. i.) and they are alt. Zs, .. AB is | CD.
(i. 27) Again : LS BGH, GHD together = two rt. Ls, (hyp.) and that LS AGH, BGH also together = two rt. ZS;
(i. 13) .. LS AGH, BGH = LS BGH, GHD; take away
the common _BGH, and the rem. _AGH = rem. LGHD,
(ax. 3) and they are alt. ZS; .. AB || CD. (i. 27) Therefore, if a straight line, &c.
Q. E. D.
1. If the opposite angles of a quadrilateral are equal to one another, the figure is a parallelogram.
2. From any point D on the base of an isosceles triangle, perpendiculars are drawn to the other two (the equal) sides : prove that the sum of the perpendiculars is the same whatever be the position of D.
PROPOSITION XXIX. THEOREM.
If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.
Let the st. line EF, fall on the || st. lines AB, CD.
Then the alt. LS AGH, GHD shall be = one other; the ext. LEGB shall be = the int. and opp. LGHD on the same side of the line EF; and the two int. Ls BGH, GHD on the same side of
EF shall together be = two rt. Zs.
Dem. For if LAGH be not=alt. LGHD, one of them
must be greater than the other; If possible, let AGH be greater than GHD; add to each
of these unequals BGH; .: LS AGH, BGH are greater than Ls BGH, GHD;
(ax. 4) But ZS AGH, BGH = two rt. Ls (i. 13); .. Zs BGH,
GHD are less than two rt. Ls. But if a st. line meets two st. lines, so as to make the
two int. Ls on the same side of it, taken together, less than two rt. Zs; these st. lines, being continually produced, shall at length meet on that side on which are the 4s which are less than two rt. ls. (ax. 12)
.. AB, CD, continually produced, will meet towards B,D;
but they never meet, since by hypothesis they are li ; :: LAGH is not unequal to LGHD, that is
LAGH = alt. LGHD. But LAGH = _ EGB; (i. 15) .. _ EGB = LGHD:
(ax. i.) add to each _ BGH; .. Ls EGB, BGH= Ls BGH, GHD; (ax. 2) but EGB, BGH=two rt. LS; (i. 13) .. also
BGH, GHD = two rt. Ls. (ax. 1)
Therefore, if a st. line, &c. Q. E. D. [This is the converse of props. 27 and 28. There is some difficulty in the theory of parallel lines, because as the definition is of a negative character (that they never meet), it is necessary to assume some positive quality on which reasonings may be founded, and this Euclid has done in ax. 12, which he regards as a self-evident theorem. Thirty methods have been proposed as starting points (for a starting point there must be), but on the whole Euclid's axiom is the simplest and least liable to objection.]
If the straight line bisecting the exterior angle of a triangle be parallel to the base, the triangle is isosceles.
PROPOSITION XXX. THEOREM.
Straight lines which are parallel to the same straight line are parallel to one another.
Let AB, CD be each of them || EF.
Constr. Let st. line GHK cut AB, EF, CD.
.. LAGH = alt. LGHF. (i. 29) Again :: GHK cuts || st. lines EF, CD at H, K;
.. ext. LGHF= int. <HKD; (i. 29) and it was shown that LAGH = LGHF;
.. LAGH = LGKD; and these are alt. Ls, :: AB is || CD. (i. 27) Therefore, straight lines which are parallel, &c.
Q. E. D.