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PROPOSITION I. PROBLEM.
Let AB be the given straight line.
From the center A, at the distance AB, describe the circle BCD;
(post. 3.) from the center B, at the distance BA, describe the circle ACE; and from C, one of the points in which the circles cut one another, draw the straight lines CA, CB to the points A, B. (post. 1.)
Then ABC shall be an equilateral triangle.
therefore AC is equal to AB; (def. 15.)
therefore B C is equal to AB;
therefore AC, BC are each of them equal to AB; but things which are equal to the same thing are equal to one another;
therefore AC is equal to BC; (ax. 1.) wherefore AB, BC, CA are equal to one another:
and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB.
Which was required to be done. PROPOSITION II. PROBLEM. 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.
From the point A to B draw the straight line AB; (post. 1.)
upon AB describe the equilateral triangle ABD, "(1. 1.) and produce the straight lines DA, DB to E and F; (post. 2.) from the center B, at the distance BC, describe the circle CGH,
(post. 3.) cutting DF in the point G: and from the center D, at the distance DG, describe the circle GKL,
cutting AE in the point L.
Then the straight line AL shall be equal to BC.
therefore BC is equal to BG; (def. 15.)
therefore DL is equal to DG,
and DA, DB parts of them are equal; (1. 1.) therefore the remainder ÂL is equal to the remainder BG; (ax. 3.)
but it has been shewn that B C is equal to BG,
wherefore AL and B C are each of them equal to BG; and things that are equal to the same thing are equal to one another ;
therefore the straight line AL is equal to BC. (ax. 1.) 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.
PROPOSITION III. PROBLEM. 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, of which A. B is the greater.
It is required to cut off from A B the greater, a part equalto C, the less.
From the point A draw the straight line AD equal to C; (1. 2.) and from the center A, at the distance AD, describe the circle DEF (post. 3.) cutting AB in the point E.
Then AE shall be equal to C.
therefore AE is equal to AD; (def. 15.)
wherefore the straight line AE is equal to C. (ax. 1.) And therefore 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.
PROPOSITION IV. THEOREM. 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 each other; 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 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, and the included angle BAC equal to the included angle EDF.
Then shall the base BC 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 angle DEF, and the angle ACB to the angle DFE.
F For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB on DE; then the point B shall coincide with the point E,
because AB is equal to DE;
and AB coinciding with DE,
the straight line AC shall fall on DF,
because AC is equal to DF;
because the point B coinciding with E, and C with F, if the base B C do not coincide with the base EF, the two straight lines BC and EF would enclose a space, which is impossible. (ax. 10.) Therefore the base B C does coincide with EF, and is equal to it;
and the whole triangle ABC coincides with the whole triangle DEF, and is equal to it; also the remaining angles of one triangle coincide with the remaining angles of the other, and are 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, &c. Which was to be demonstrated.
PROPOSITION V. THEOREM. The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal. Let ABC be an isosceles triangle of which the side AB is equal to AC,
and let the equal sides AB, AC be produced to D and E.
In BD take any point F;
and join FC, GB. Because AF is equal to AG, (constr.) and AB to AC; (hyp.) 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
therefore the base FC is equal to the base GB, (I. 4.)
and the triangle AFC is equal to the triangle AGB, also the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite;
viz. the angle ACF to the angle ABG,
and the angle AFC to the angle AGB.
of which the parts AB, AC, are equal; therefore the remainder BF is equal to the remainder CG; (ax. 3.)
and FC has been proved to be equal to GB; hence, because the two sides BF, FC are equal to the two CG, GB,
each to each ; and the angle BFC has been proved to be equal to the angle CGB, also the base BC is common to the two triangles BFC, CGB;
wherefore these triangles are equal, (1. 4.) 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; therefore the remaining angle AB Cis equal to theremaining angle ACB,
which are the angles at the base of the triangle ÅBČ;
that the angle FB Ĉ is equal to the angle GCB,
Therefore the angles at the base, &c. Q.E.D.
PROPOSITION VI. THEOREM.
Then the side AB shall be equal to the side AC.
For, if AB be not equal to AC, one of them is greater than the other.
If possible, let AB be greater than AC; and from BA cut off BD equal to CA the less, (1. 3.) and join DC.
Then, in the triangles DBC, ABC, because DB is equal to AC, and BC is common to both triangles, the two sides DB, B Care equal to the two sides AC, CB, each to each ;
and the angle DBC is equal to the angle ACB; (hyp.)
the less equal to the greater, which is absurd. (ax. 9.) Therefore AB is not unequal to A C, that is, AB is equal to AC.
Wherefore, if two angles, &c. Q.E.D.
PROPOSITION VII. THEOREM. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremity.
If it be possible, on the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their sides CB, DB, that are terminated in B.
Join CD. First. When the vertex of each of the triangles is without the other triangle.
Because AC is equal to AD in the triangle ACD, therefore the angle ADC is equal to the angle ACD; (1. 5.) but the angle ACD is greater than the angle BCD; (ax. 9.)
therefore also the angle ADC is greater than BCD; much more therefore is the angle BVC greater than BCD. Again, because the side B C is equal to BD in the triangle BCD, (hyp.)
therefore the angle BDC is equal to the angle BCD; (1. 5.)
but the angle BDC was proved greater than the angle BCD, hence the angle BDC is both equal to, and greater than the angle BCD;
which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle ACB.