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straight line; where the point is not situate K
in any part of the given line. It is required H
to draw from the point A a straight line equal D A
to BC.

B В

G
*INTERC. 9. From the point A to B draw* the straight line
Cor.
+1. 1.

AB, and upon it describet the equilateral triangle
INTERC. 12. DAB, and prolong the straight lines DA and DB, to E and F;
Cor. 6.
*INTERC.13.

which will* be in the same plane with the rest. About the centre
Cor. 2. B, with the radius BC, describet the circle CGH; and about the
+INTERC.13.
Cor. 3.

centre D, with the radius DG, describe the circle GLK, cutting Interc.13. DE in the point L. AL shall be equal to BC. Cor. 5. *INTERC,13.

Because the point B is the centre of the circle CGH, BC is* Cor. 4. equal to BG; and because D is the centre of the circle GLK, + Constr. DL is equal to DG. And DA, DB, parts of them, aret equal ; Interc. 1. therefore the remainder AL ist equal to the remainder BG. Cor. 7.

But it has been shown that BC is equal to BG; wherefore AL

and BC are each of them equal to BG. And things which are *Interc. 1. equal to the same, are* equal to one another; therefore AL is equal to BC.

Wherefore from the point A a straight line AL has
been drawn, equal to BC.
Second Case; where A the point assigned is

K
situate in some part of the given straight line BC,
but not in one of its extremities.

Upon either of the portions AB, AC, (as, for

instance, upon AB), describet an equilateral + I. 1.

F
Interc12. triangle DAB, and prolongf the straight lines DA and DB, to E
Cor. 6.
*Interc.1?, and F. About the centre B, with the radius BC, describe* the
Cor. 3.

circle CGH; and about the centre D, with the radius DG, describe
the circle GLK, cutting DE in L. AL shall be equal to BC.

Because the point B is the centre of the circle CGH, 8c. [the
remainder of the proof is word for word as in the last Case.]

H
D
B

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Third Case; where the point assigned is situate in one of the extremities of the given straight line BC;

as, for instance, in B. INTERC.13. About the centre B, with the radius BC, describet Cor. 3.

the circle CLH.

If from B any straight line, as BE, be drawn to a point without the circle, and L be a point where it is cut by the circumference, then because the point B is the centre of the circle CLH, and

*Interc.13. L and C are points in the circumference, BL is* equal to BC. Cor. 4.

Wherefore from the point B: a straight line has been drawn, equal to BC.

In all the possible Cases, therefore, has been shown how from an assigned point to draw a straight line equal to a given straight line BC. Which was to be done. And by parity of reasoning, the like

may be done in every other instance. ScHolium.—It has not yet been proved, that there might not be more equilateral triangles than one, described upon AB on the same side, and that the prolongations of these might not make straight lines from B other than BF. Which if it were possible (though it will hereafter be shown that it is not), would in no way affect the accuracy of the assertion that it has been shown how from A to cut off a straight line equal to BC.--Referred back to, in the Scholium at the end of Prop. VII of the First Book.

PROPOSITION III.

PROBLEM.From the greater of two given straight lines, to cut

off a part equal to the less.

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

A

JEB From either of the ends of AB (as, for + I. 2. instance, from A), drawt a straight line AD equal to C; [which

straight line will fall in some direction that will be determined by the operations described in the preceding Problem, and will not, except by accident, coincide with the given straight line

AB, from which a part equal to C is desired to be cut off']. INTERC.13. About the centre A, with the radius AD, describe the circle Cor. 3. *Interc.13. DEF, cutting AB in the* point E. Cor.5. Because A is the centre of the circle DEF, and E and D are +Interc.13,

points in the circumference, AE ist equal to AD. But the Cor. 4.

straight line C is likewise equal to AD; whence AE and C are INTERC. 1. each of them equal to AD. Wherefore AE ist equal to C, and

from AB the greater of the two straight lines, a part AE has been cut off, equal to C the less. Which was to be done.

And by parity of reasoning, the like may be done in every other instance.

PROPOSITION IV.

D

B

C

Cor. 3.

THEOREM.-If two triangles have two sides of the one, equal

to two sides of the other respectively; and have also the

angles belween those sides, equal to one another ; they shall See Note, have their third sides equal ; and the two triangles shall be

equal; and their other angles shall be equal respectively, viz. those to which equal sides are opposite.

Let BAC, EDF be two triangles, which have the two sides AB, AC of the one, equal to the two sides DE, DF of the other respectively, viz. AB to DE, and AC to DF;

F and the angle 'BAC equal to the angle EDF. The third side BC shall be equal to the third side EF, and the triangle BAC to the triangle EDF; and the other angles, to which the equal sides are opposite, shall be equal respectively, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

For, if the triangle BAC was applied to the triangle EDF, so INTERC.12.

that the point A was on D, and the straight line AB on*

DE; the point B would coincide with the point E, because AB + Hyp.

ist equal to DE. And, AB so coinciding with DE, if the triangle BAC was turned round the straight line AB till the point

C was made to be in the plane in which is the triangle DEF, the INTERc.14.

planes in which are the two triangles would coincide throughout. And because the angle BAC is equal to the angle EDF, the straight line AC would coincide with the straight line DF, to the extent of the length common to both ; for if it did not, the angle BAC

would be either greater or less than EDF, which is impossible, + Hyp. for they aret equal. And because the straight line AC is equal Hyp.

to DF, the point C would coincide with the point F. But the

point B coincides with the point E; wherefore (the point B coin* INTERC,12.

ciding with E, and C with F,) the straight line BC would* coincide 11. Nom.14. with the straight line EF; and coinciding, they would be equalt.

Wherefore the whole triangle BAC would coincide with the whole triangle EDF, and be equal to it; and the remaining angles of the one would coincide with the remaining angles of the other respectively, and be equal to them, yiz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

Cor. 3.

• Hyp:

Cor. 1.

And by parity of reasoning, the like may be proved of all other triangles under the same conditions ; that is to say, which have two sides of the one equal to two sides of the other respectively, and also the angles between those sides equal to one another. Wherefore, universally, if two triangles have two sides of the one, equal to two sides of the other respectively ; &c. Which was to be demonstrated.

* Hyp:

F

PROPOSITION V.
THEOREM.-In any isoskeles triangle, the angles opposite to two

equal sides are equal to one another.
Let BAC be an isoskeles triangle, of which the side AB is equal
to AC.

The angles ACB and ABC shall be equal to one another. *INTERC.12. Prolong* AB and AC to any lengths, as for instance to D and Cor. 6. + 1. 3.

E. In BD take any point F, and from AE cut off† AG equal to

AF, and join BG, CF.
# Constr. In the triangles FAC, GAB, because AF ist

A
equal to AG, and AC to* AB, the two sides
AF, AC are equal to the two AG, AB re-
spectively; and the angle at A is common to

B В

C the two triangles FAC, GAB; therefore the + 1. 4. third side FC ist equal to the third side GB,

G and the triangle FAC to the triangle GAB;

D

E and the remaining angles of the one are equal to the remaining angles of the other respectively, to which the equal sides are opposite; viz. the angle AFC to the angle AGB, and the angle ACF to the angle ABG. Also, because the whole AF is equal to the

whole AG, of which the parts AB, AC are equal, the remainder Interc. 1. BF will bef equal to the remainder CG. And FC was shown to Cor.7.

be equal to GB; therefore in the triangles BFC, CGB, the two sides FB, FC are equal to the two GC, GB respectively; and

moreover the angle BFC was shown to be equal to the angle CGB; • I. 4.

wherefore the triangles are* equal, and the remaining angles of the one are equal to the remaining angles of the other respectively, viz. those to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to CBG. And since it has been demonstrated that the whole angle ACF is equal to the whole angle ABG, parts of which, the angles

BCF and CBG, are also equal; it follows that the remainders * INTERC. 1. are* equal, viz. the angle ACB equal to the angle ABC. Cor. 7.

And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, in an isoskeles triangle the

angles opposite to two equal sides are equal to one another. See Note.

Which was to be demonstrated.

Cor. 1. If two equal sides of any isoskeles triangle be prolonged, the two angles exterior to the third side shall be equal.

For the angle FBC was shown to be equal to GCB.
Cor. 2. Every equilateral triangle is also equiangular.

For if ABC be an equilateral triangle, then because
the sides CB and CA are equal, the angle A
(by Prop. V above) is equal to the angle B.
Again, because the sides AB and AC are equal, the А

B angle C is equal to the angle B. But the angle A also was *Interc. 1. equal to the angle B; therefore the angles A and C aret equal

to one another. And since the angles A and C are also equal to B, the angles A, B, C are all equal to one another.

PROPOSITION VI.

| 1.3.

THEOREM.If two angles of a triangle are equal to one another,

the sides opposite to the two equal angles are equal to one another.
Let BAC be a triangle having the angle ACB

D.
equal to the angle ABC; the side BA is equal to
the side CA.

For if BA be not equal to CA, one of them
must be greater than the other. Let BA be B

с assumed to be the greater, and from it cut off# BD equal to CA, and join CD.

Because in the triangles DBC and ACB, B Dis equal to CA, and

BC common to both, the two sides BD and BC must be equal to the • Hyp.

two CA, CB respectively. And the angle DBC is* equal to the + I. 4.

angle ACB. Therefore the triangle DBC must bet equal to

the triangle ACB, the less to the greater; which is impossible. 11. Nom.26. The assumptiont, therefore, which involves this impossible con

sequence, cannot be true; or BA is not greater than CA. And in the same way may be shown, that CA is not greater than

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