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 ▲ 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. Hence

(Ax. 10)

ABC would occupy the former position of ▲ ACB, and DBC the former position of ▲ ECB.

·: ▲ ABC = 2 ACB and ▲ 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 AABC, and the same ▲ ACB.

.. the s opposite BA and CA are equal.

3. PROPOSITION VII.-THEOREM.

(I. 4)

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 ▲ lies without the other A, take E in AC and F where AD cuts BC;

when the vertex of one ▲ lies within the other,

take E in AC produced and F where AD produced cuts BC.

The case in which the vertex of one ▲ 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.

Let BA make with CD the adj. 4s CBA, DBA.
Then CBA and DBA shall together =two rt. LS.
ΑΙ

(Def. 10)

(I. 11)

For, if CBA =

DBA, each is right.

But, if CBA does not = DBA,

draw BE LCD, so that CBE, EBD are rt. Ls.

Then ▲ DBA and a part of ABC together make up the rt. DBE,

and the remaining part of ABC forms the other rt. ▲ EBC. Therefore the two 2 s DBA, ABC together = two rt. 4 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 ▲ ;

then shall AC, CB be > AB; CB, BA be > AC; and BA, AC > BC.

Bisect the ACB by the str. line CD, meeting AB in D,

A

D

B

(I. 9)

Then ACD = ▲ BCD; but ▲ ADC is > BCD.

Similarly,

..

▲ ADC is > ▲ ACD;

.. side AC is > AD

side BC is > BD.

.. the two sides AC, CB together are > AB.

(I. 16)

(I. 19)

In the same way we may prove the other inequalities.
Hence, any two sides, &c.

6. PROPOSITION XXIV.-THEOREM.

Q. E. D.

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 > LEDF.

the base EF.

Then shall the base BC be
Make CAG = ▲ FDE (I. 23), AG

= DE; and join CG. Then CA, AG = FD, DE, each to each; and 4 CAG=FDE;

..the base CG base EF.

(I. 4)

:: ▲ BAC is > half of BAG,

the bisecting line will meet BC.

Let it meet BC in H; and join GH.

F

(I. 9)

=

Then BA, AH= GA, AH, each to each, and ▲ BAH=

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

LB = LE, 0= ▲ F, and side BC = side EF. Then shall AB = DE, AC = DF, and ▲ A = 2 D.

L

Apply ▲ ABC to ADEF, putting B on E and BC on EF.

.. BA will coincide with ED and be equal to it, CA will coincide with FD and be equal to it, and 4 A will coincide with 2D and be equal to it. Secondly, let the As ABC, DEF have

(Ax. 8)

LB = LE, 2 C = ▲ DFE, and side AB = side DE. In this case, also, they shall be equal in all respects.

Apply AABC to ADEF, putting A on D and AB on DE.

For, if not, let it meet EF or EF produced in H.

Then, in either case,

: 40 = 4DFE ; ·

L DHE = 4 DFE;

which is impossible.

and, as before, the As ABC, DEF coincide;

and are equal in all respects.

(I. 16)

.. AC does fall on DF;

(Ax. 8)

Wherefore, &c.

Q. E. D.