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For, if the complete figure were taken up, reversed, and then placed with A in its former position,
and AD along the former position of AE; then, :; DAE and 2 EAD are the same,
AE would fall along the former position of AD. Also,
.:: AB = AC, B would fall where C was, and C would fall where B was; ::. BC would fall where CB was.
(Ax. 10) Hence ABC would occupy the former position of Z ACB, and < DBC the former position of ZECB. ::: Z ABC = LACB and Z DBC = ECB. (Ax. 8)
Q. E. D.
Note 1.-A proof similar in nature to the above is easily constructed for Prop. VI.; but no saving of space is thereby effected.
Note 2.—The first part of Prop. V. may be proved as a particular case of Prop. IV. For BA, AC, and _BAC
CA, AB, and <CAB, each to each, in the A ABC, and the same A ACB. .. the 28 opposite B A and CA are equal.
3. PROPOSITION VII.-THEOREM. Upon the same base and upon the same side of it there cannot be two triangles which have not only their sides which are terminated in one extremity of the base equal to one another, but also those which are terminated in the other extremity. If it be possible, upon the same base, and on the same side
of it, let there be two As ACB, ADB, which have not only CA, DA terminated in A equal, but also CB, DB terminated in B.
When the vertex of each A lies without the other A, take E in AC and F where AD cuts BO; when the vertex of one A lies within the other, take E in AC produced and F where AD produced cuts BC.
::: AC= AD,
(I. 5) .. the ZFDC is > the 2 FCD. Again,
.: BC= BD,
(I. 5) :. the LFDC is < the ZFCD; so that the _ FDC is both > and < the ZFCD; which is absurd. The case in which the vertex of onė A lies upon a side of
the other needs no proof. Wherefore, &c.
Q. E. D. Note. - The proof just given will be seen to apply at the same time to both the cases usually given.
4. PROPOSITION XIII.—THEOREM.
Any two adjacent angles which one straight line makes with
another, are together equal to two right angles.
But, if CBA does not draw BE LCD, so that CBE, EBD are rt. Z 8.
(I. 11) Then _ DBA and a part of _ ABC together make up
the rt. ZDBE, and the remaining part of ABC forms the other rt. EBC. Therefore the two Zs DBA, ABC together = two rt. Z s.
Q. E. D. Note.—A considerable simplification may be introduced into the usual proof, as already given in the text, by merely denoting the three small angles of the second figure by a, b, y. The same expedient may be used with advantage in XIV., XV., XXVIII., XXIX.
5. PROPOSITION XX.—THEOREM. Any two sides of a triangle are together greater than the third side.
Let ABC be any A ; then shall AC, CB be > AB; CB, BA be > AC; and BA,
AC > BC. Bisect the ZACB by the str. line CD, meeting AB in D,
Then LACD = LBCD; but _ADC is > BCD. (I. 16)
.. LADC is > LACD;
(I. 19) Similarly, side BC is > BD. .. the two sides AC, CB together are > AB.
In the same way we may prove the other inequalities.
Q. E. D.
6. PROPOSITION XXIV.-THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, but the contained angle of the one greater
than the contained angle of the other, then the base of that which has the greater angle shall be greater than the base of the other.
In the As ABC, DEF, let BA, AC = ED, DF, each to each, but let the <BAC be > ZEDF.
Then shall the base BC be > the base EF.
Make ZCAG = LFDE (I. 23), AG = DE; and join CG. Then CA, AG = FD, DE, each to each; and 2 CAG=FDE; .:. the base CG = base EF.
(I. 9) ::: ZBAC is > half of ZBAG, the bisecting line will meet BC. Let it meet BC in H; and join GH. Then BA, AH = GA, AH, each to each, and BAH = ZGAH;
.. the base BH = base GH,
:. the base BC= GH, HC together ; But GH, HC together are > GC.
(I. 20) .:. BC is > GC; i. BC is > EF. Q.E.D.
7. PROPOSITION XXVI.-THEOREM.
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, namely, either the sides adjacent to the equal angles or the sides opposite to either pair of equal angles; then shall their other sides be equal, each to each, and also their third angles. First, let the As ABC, DEF have
ZB = LE, 2C= _F, and side BC = side EF. Then shall AB = DE, AC = DF, and LA = D.
Apply A ABC to ADEF, putting B on E and BC on EF. Then C will fall on F, :: BC= EF; Also BA will fall on ED, ::: ZB :
; and CA will fall on FD, .:: ZC= _F.
.. A will fall on D; :. BA will coincide with ED and be equal to it, (Ax. 8)
CA will coincide with FD and be equal to it, and ZA will coincide with 2D and be equal to it. Secondly, let the As ABC, DEF have
ZB = LE, 2C = ZDFE, and side AB = side DE. In this case, also, they shall be equal in all respects. A
H F E
F H Apply A ABC to A DEF, putting A on D and AB on DE.
Then B will fall on E, ::: AB = DE; also BC will fall on EF, ::: ZB = LE; also AC will fall on DF.
For, if not, let it meet EF or EF produced in H.
::20 = ZDFE; .. ZDHE = DFE ; which is impossible.
(I. 16) .. AC does fall on DF; and, as before, the As ABC, DEF coincide; and are equal in all respects.
(Ax. 8) Wherefore, &c.
Q. E. D.