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PROPOSITION I. PROBLEM.
UPON a given finite right line to describe an equilateral triangle.
Let AB be the given right line ;- it is required to describe an equilateral triangle upon it,
From the point A, at the distance AB, describe the circle BCD (Pos. 3.)
And from the point B, at the distance BA, describe the circle Ace (Pol. 3.)
Then, because the two circles pass through each other's centres, they will cut each other.
And, if the right lines CA, CB be drawn from the point of intersection c, Abe will be the equilateral triangle required.
For, fince A is the centre of the circle BCD, AC is equal to AB (Def. 13.)
And, because B is the centre of the circle ACE, Bc is also equal to AB (Def. 13.)
But things which are equal to the same thing are equal to each other (Ax. 1); therefore Ac is equal to CB.
And, since AC, ce are equal to each other, as well as to AB, the triangle, ABC is equilateral; and it is described upon the right line AB, as was to be done.
PRO P. II. PROBLEM.
From a given point to draw a right line equal to a given finite right line.
Let A be the given point, and bc the given right line ; it is required to draw a right line from the point A, that thall be equal to BC.
Join the points A, B, (Pof. 1.); and upon ba describe the equilateral triangle BAD (Prop. 1.)
From the point B, at the distance Bc, describe the cir. cle cer (Pos. 3.) cutting DB produced in F.
And from the point d, at the distance DF, describe the circle Fig (Pos. 3.), cutting DA produced in G, and AG will be equal to BC, as was required.
For, since B is the centre of the circle CEF, Bc is equal to bf (Def. 13.)
And, because D is the centre of the circle FHG, DG is equal to DF (Def. 13.)
But the part da is also equal to the part DB (Def. 16.), whence the remainder AG will be equal to the remainder BF (Ax. 3.)
And since AG, Bc have been each proved to be equal to BF, AG will also be equal to BC (Ax. 1.)
A right line Ag, has, therefore, been drawn from the point A, equal to the right line BC, as was to be done.
SCHOLIUM. When the point A is at one of the extremities B, of the given line Bc, any right line, drawn from that point to the circumference of the circle CEF, will be the one required.
PRO P. III. PROBLEM.
From the greater of two given right lines, to cut off a part equal to the less.
Let AB and c be the two given right lines; it is required to cut off a part from AB, the greater, equal to c the less.
From the point a draw the right line Ad equal to c (Prop. 2.); and from the centre A, at the distance AD, describe the circle der (Pol. 3.) cutting AB in E, and AE will be equal to c as was required.
For, since A is the centre of the circle EDF, AE will be equal to AD (Def. 13.)
But c is equal to AD, by construction; therefore AE will also be equal to c (Ax. 1.)
Whence, from AB, the greater of the two given lines, there has been taken a part equal to c the less, which was to be done.
SCHOLIUM. When the two given lines are so situated, that one of the extremities of c falls in the point A, the former part of the construction becomes unnecessary,
PRO P. IV. THEOREM.
If two sides and the included angle of one triangle, be equal to two fides and the included angle of another, each to each, the triangles will be equal in all respects.
Let ABC, der be two triangles, having cA equal to FD, CB to Fe, and the angle c to the angle F; then will the two triangles be equal in all respects.
For conceive the triangle ABC to be applied to the triangle DEF, so that the point c may coincide with the point F, and the side CA with the side FD.
Then, because CA coincides with FD, and the angle c is equal to the angle F (by Hyp.), the fide CB will also coincide with the side Fe.
And, since ca is equal to FD, and CB to Fe (by Hyp.), the point A will fall upon the point D, and the point B upon the point E.
But right lines, which have the same extremities, muft coincide, or otherwise their parts would not lie in the fame direction, which is absurd (Def. 5.); therefore AB falls upon, and is equal to de.
And, because the triangle ABC is coincident with the triangle DEF, the angle A will be equal to the angle D, the angle B to the angle E, and the two triangles will be equal in all respects (Ax. 9.) Q. E. D.
PROP. V. THEOREM.
The angles at the base of an isosceles triangle are equal to each other.
Let ABC be an isosceles triangle, having the fide ca equal to the fide CB; then will the angle CAB be equal to the angle CBA.
For, in ca and ce produced, take any two equal parts CD, ce (Prop. 3), and join the points AE, BD:
Then, because the two fides CA, CE of the triangle CAE, are equal to the two sides CB, CD of the triangle CBD, and the angle c is common, the fide Ae will also be equal to the side BD, the angle CAE to the angle CBD, and the angle to the angle E (Prop. 4.)
And since the whole cd is equal to the whole ce (by Conft.), and the part ca to the part CB (by Hyp.), the remaining part AD will also be equal to the remaining part Be (Ax. 3.)
The two sides DA, DB, of the triangle DAB, being, therefore, equal to the two sides EB, EA of the triangle EBA, and the angle o to the angle e, the angle ABD will also be equal to the angle BAE (Prop. 4.)
And if from the equal angles CAE, CBD, there be taken the equal angles BAE, ABD, the remaining angle