Post. ).

Draw AB; on AB descr. an equilat. A Prop. 1. DAB; prod. DA, DB to E and F; from cr. B

and dist. BC, descr. O CGH; from cr. D and Post. 3. dist. DG, descr. GKL. Then AL - BC.

Post. 2.

Def. 15.

Def. 15.
Constr.
Ax. 3.

B is cr. O CGH,

BC = BG;
.:: D is cr. O GKL,

DL = DG;
and part DA = part DB;
.. rem. AL = rem. BG;
but BC = BG;

... AL = BC.
Wherefore, from the given pt. A, a str. line
AL has been drawn equal to the given str.
line BC.

Ax. 1.

From the greater of two given straight lines,

to cut off a part equal to the less. Let AB and C be the given str. Jines, whereof AB > C; it is required to cut off from AB a part = C.

From A draw AD = C; from cr. A and Prop. %. dist. AD, descr. O DEF, cutting AB in E ; Post. 3. then AE = C.

:: A is cr. O DEF, i. AE = AD; but C= AD;

.. AE Wherefore, from AB, the greater of two str. lines, a part AE has been cut off = C, the less.

Def. 15. Constr. Ax. 1.

= C.

PROP. IV. THEOR.

4. 1 Eu.

If two triangles have two sides of the one, equal

to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another; they shall likewise have their bases, or third sides, equal, and the two triangles shall be equal ; and their other angles shall be equal, each to each, viz., those to which the equal sides are opposite.

Let ABC, DEF, be two As, of which AB =DE, AC DF, and < BAC = Z EDF. Then BC EF, A ABC = A DEF, Z ABC = Z DEF, and < ACB = ZDFE.

Нур.

AB =

DE,

For if ~ ABC be applied to A DEF, so that pt. A may be on D, and AB on DE.

.::

B coincides with E.

AB concides with DE, Нур.

BAC = L EDF;

AC falls upon DF. Нур. . And

AC = DF,

C coincides with F.
But

B coincides with E,

BC coincides with EF. For if BC do not coincide with EF, then two Ax. 10. str. lines enclose a space, which is impossible.

BC coincides with and

ABC coincides with and => DEF,
LABC coincides with and = _DEF,

L ACB coincides with and = _DFE.
Therefore, if iwo triangles have, &c.

EF,

PROP. V. THEOR. 5. I Eu.
The angles at the base of an isosceles triangle

are equal to one another; and if the equal
sides be produced, the angles upon the other
side of the base shall be equal.
Let ABC be an isosc. A, having AB = AC.

Let AB and AC be prod. to D and E. Then
L ABC = L ACB, and % CBD = _ BCE.

Prop. 4.

Constr.

In BD take any pt. F; from AE, the greater, Prop. 3. cut off AG=AF; join FC, GB.

Constr.
ŞAF = AG,
AB = AC,

Нур. and / FAG common to As AFC, AGB,

base FC = base GB,
L ACF = L ABG,

2 AFC = L AGB. Again, :: whole AF = whole AG, and part AB = part AC,

Нур. : .. remain. BF = remain. CG;

FC = GB,

BF = CG,
L BFC = L CGB;

BCF =
...
L FBC =
SGCB, the son op-

Prop. 4.

posite side of base. But whole L ABG = whole ACF, and part 2 CBG = part _ BCF; .. rem. _ ABC =

Įrem. L ACB, the

Ls at the base.

Ax. 3.

and •:

% CBG

Ax. 3.

:. Therefore, the angles, &c.

CoR.-Hence every equilateral is also equiangular.

PROP. V. THEOR.

OTHERWISE DEMONSTRATED.

The angles at the base of an isosceles triangle

are equal to one another. Let ABC be an isosc. A, having the side AB = side AC; then will % ABC = L ACB.

A

Нур. .

B DC Let the str. line AD divide the L BAC into two = parts. (AD is common to both As ABD, ACD,

and AB = AC,

_BAD = L DAC;

A ABD = A ACD, Prop. 4. ..

at the base. Wherefore the angles at the base, &c.

Note. It is evident that some line, as AD, will bisect the < BAC; and although the method of bisection is not known until Pro. 8 be solved, yet this does not affect the truth of