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Book I.

X.
Two straight lines cannot enclose a space.

XI.
All right angles are equal to one another.

XII.
“ 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 straight lines being continually
“ produced, shall at length meet upon that side on which are
« The angles which are less than two right angles. See the
“ notes on prop. 29. of Book I.".

4

PROPOSITION I. PROBLEM.

Book I.

TO describe an equilateral triangle upon a given finite straight line.

Let AB be the given straight line ; it is required to describe an equilateral triangle upon it.

с From the centre A, at the distance AB, describe a the circle

& 3. PostuBCD, and from the centre B, at

late. the distance BA, describe the

A circle ACE; and from the point

B E C, in which the circles cut one another,draw the straight lines b

b 2. Post. CA, CB to the points A, B; ABC shall be an equilateral triangle.

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Because the point A is the centre of the circle BCD, AC is cqual e to AB ; and because the point B is the centre of the cir. c 15. Deficle ACE, BC is equal to BA: but it has been proved that CA is nition. equal to AB; therefore CA, CB are each of them equal to AB; but things which are equal to the same are equal to one another; d therefore CA is equal to CB; wherefore CA, AB, BC are

d 1st Axiequal to one another; and the triangle ABC is therefore equila- om. teral, and it is described upon the given straight line AB. Which was required to be done.

PROP. II. PROB.

FROM a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC,

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Book I.

Because the point B is the centre of the circle CGH, BC is

equal e to BG; and because D is the centre of the circle GKL, e 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal ; f 3. Ax. therefore the remainder AL is equal to the remainder f BG: but

it has been shown, that BC is equal to BG; wherefore AL and
BC are each of them equal to BG; and things that are equal to
the same are equal to one another; therefore the straight line
AL is equal to BC. Wherefore from the given point A a straight
line AL has been drawn equal to the given straight line BC.
Which was to be done.

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PROP. III. PROB.

t + FROM the greater of two given straight lines to cut off a part equal to the less.

Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less.

А a 2. 1. From the point A draw a the

straight line AD equal to C; and

from the centre A, and at the disb 3. Post, tance AD, describe b the circle DEF;and hecause A is the cen

F
tre of the circle DEF, AE shall be equal to AD; but the straight
line C is likewise equal to AD; whence AE and C are each of

them equal to AD; wherefore the straight line AE is equal to 1. AX. c C, and from AB, the greater of two straight lines, a part AE

has been cut off equal to C the less. Which was to be done.

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PROP. IV: THEOREM.

IF two triangles have two sides of the one equal
to two sides of the other, each to each; and have likc.
wise the angles contained by those sides equal to one
another; they shall likewise bave 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 triangles which have the two sides
AB, AC equal to the two sides DE, DF, each to each, viz.

"AB to DE, and AC to DF; A

D

Book I. and the angle BAC equal to the angle ÉDF, the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles, to which the equal sides are opposite, shall be equal each to each, viz. the angle ABC to the B

с Е angle DEF, and the angle ACB to DFE.

For, if the triangle ABC be applied to DEF, so that the poi A may be on D, and the straight line AB upon DE; the point B shall coincide with the point. E, because AB is equal to DE ; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF: but the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impossible. Therefore a 10. Åx. the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, 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, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was. to be demonstrated.

PROP. V. THEOR.

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 isosceles triangle, of which the side AB is

с

b

Book I. equal to AC, and let the straight lines AB, AC be produced to
SD and E, the angle ABC shall be equal to the angle ACB, and

the angle CBD to the angle BCE.

In BD take any point F, and from AE the greater, cut off a 3. 1. AG equal to AF, the less, and join FC, GB.

Because AF is equal to AG, and AB to AC, the two sides
FA, AC are equal to the two GA, AB, each to each; and they
contain the angle FAG common
to the two triangles AFC, AGB; as
therefore the base FC is equal
to the base GB, and the triangle
AFC to the triangle AGB ; and
the remaining angles of the one pie
jare equal to the remaining an- ja

C
gles of the other, each to each,
to which the equal sides are op-
sposite; viz. the angle ACF to
the angleABG, and the angle
AFC to the angle AGB: and
because the whole AF is equal D

E
ito the wholé AG, of which the

parts AB, AC, are equal ; the Auto renainder BF shall be equal to the remainder CG ; and FC

was proved to be equal to GB ; therefore the two sides BF, FC are equal to the two CG, GB, each to each : and the angle BFC is equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal, and their remaining angles, each to each, to which the equal sides are opposite ; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG: and, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG,

BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: and it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the

gles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equi. angular.

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IF two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

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