THEOREM ().

Upon the same base and upon the same side of it there cannot be two triangles which have their conterminous sides equal.

For if it be possible upon the same base BC and on the same side of it let there be two as ABC, DBC which have their conterminous sides equal, viz. AB = DB__and AC DC.

=

Ist.

Let the vertex of each a fall without the other a.

but BAD is > L CAD; .. L BDA is > L CAD:

but since AC = DC, it follows that CDA is = L CAD,

which is impossible.

2nd. Let the vertex of one of the as as A fall within the other ▲ DBC.

A

G

B

Join AD; and produce BD, BA to F, G.

but FDAL CDA; .. 4 GAD> L CDA;

but ▲ CAD> ▲ GAD;

3rd. Let the vertex A of one of the as fall upon one of the sides of the other.

B

Then it is evident that AC is < DC,
but by hypothesis it is = it, which is impossible.
Wherefore upon the same base, &c.

N.B. By the aid of this Theorem Prop. (1. v.) may easily be established.

Euclid's Twelfth Axiom, on which his Theory of Parallels depends, is "If a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these two straight lines being continually produced, shall at length meet on that side on which are the angles which are less than two right angles."

Hence Propositions XXI. and XXII. are established as follows:

PROPOSITION XXI.

If a straight line meets two parallel straight lines it makes the alternate angles equal to one another.

Let the straight line FG meet the two || straight lines AB, CD, then shall ▲ AFG = the alternate 4 FGD.

For, if not, one of them must be the greater.

Let AFG be greater than FGD;

.. LS AFG, GFB are together >

but 4S AFG, GFB are together

=

Ls FGD, GFB,

two rights;

.. Ls FGD, GFB are together < two rights;

.. AB, CD will meet towards B, D; but they do not.

PROPOSITION XXII.

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

Let the straight lines A, B be each of them || to C,
then shall A be || to B.

Draw OPQR cutting A, B, C in P, Q, R.

PROPOSITION XXV. COR.

If two angles of one triangle are equal to two angles of another, and any side of one is equal to the corresponding side of the other, then shall the triangles be equal in all respects.

The case in which the given sides in each are adjacent to the equal angles is established in Prop. III.

Euclid proves the case in which the given equal sides are opposite to equal angles in each thus :

A A

Let ABC, DEF be two as having ABC = ▲ DEF, LACB= = LDFE, and side AB = side DE.

Then BC and EF are equal to one another,

for, if not, one of them must be the greater,

let this be EF, and from EF cut off EX= BC; join DX.

Then . AB = DE, BC = EX, and ▲ ABC = L DEX;

.. LACB
L

but 4 ACB = 4 DFE; .. LDXE = ¿ DFE;

the exterior

of a ▲ the interior and opposite angle ;

=

which is impossible;

.. BC is = EF.

Also AB = DE and ▲ ABC = L DEF;

.. ▲ ABC is = ▲ DEF in all respects.