Sidebilder
PDF
[ocr errors][subsumed][ocr errors][merged small][ocr errors][ocr errors][merged small]

-

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, where-
of AB > C; it is required to cut off from AB
a part = C.

[ocr errors][ocr errors][ocr errors][ocr errors]

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 = LEDF. Then BC = EF, A ABC = A DEF, L ABC = Z DEF, and < ACB = 2 DFE.

[ocr errors]
[ocr errors]

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

AB = DE,
:: B coincides with E.

..J AB concides with DE, Hyp

BAC LEDF;

AC falls upon DF.
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 = EF, JA ABC coincides with and =A DEF, " LABC coincides with and = _DEF,

LACB coincides with and = LDFE. Therefore, if iwo triangles have, &c.

PROP. V. THEOR. 5. 1 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.

[ocr errors][subsumed][ocr errors][ocr errors]

:. Therefore, the angles, &c.

Cor.-Hence every equilateral A equiangular.

is also

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.

[ocr errors]

в р с
Let the str. line AD divide the L BAC
into two = parts.
SAD is common to both As ABD, ACD,

and AB = AC,
Z BAD = L DAC;

A ABD = A ACD,
Prop. 4. LAC JZ ACB, which are the Ls

12 ABO = 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

« ForrigeFortsett »