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11. An obtuse angle is that which is greater than a right angle.
12. An acute angle is that which is less than a right angle.
13. A term or boundary is the extremity of anything.
14. A figure is that which is enclosed by one or more boundaries. 15. A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another.
16. And this point is called the centre of the circle.
17. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
18. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter.
19. A segment of a circle is the figure contained by a straight line, and the circumference it cuts off.
20. Rectilineal figures are those which are contained by straight lines.
21. Trilateral figures, or triangles, by three straight lines.
22. Quadrilateral by four straight lines.
23. Multilateral figures, or polygons, by more than four straight lines.
24. Of three-sided figures, an equilateral triangle is that which has three equal sides.
25. An isosceles triangle is that which has two sides equal.
26. A scalene triangle is that which has three unequal sides. 27. A right-angled triangle is that which has a right angle.
28. An obtuse-angled triangle is that which has an obtuse angle.
29. An acute-angled triangle is that which has three acute angles. 30. Of four-sided figures, a square is that which has all its sides equal, and all its angles right angles.
31. An oblong is that which has all its angles right angles, but has not all its sides equal.
32. A rhombus is that which has all its sides equal, but its angles are not right angles.
33. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles. 34. All other four-sided figures besides these are called Trapeziums. 35. Parallel straight lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet.
1. Let it be granted that a straight line may be drawn from any one point to any other point.
2. That a terminated straight line may be produced to any length in a straight line.
3. And that a circle may be described from any centre, at any distance from that centre.
1. Things which are equal to the same are equal to one another.
2. If equals be added to equals, the wholes are equal.
3. If equals be taken from equals, the remainders are equal. 4. If equals be added to unequals, the wholes are unequal.
5. If equals be taken from unequals, the remainders are unequal. 6. Things which are double of the same are equal to one another.
7. Things which are halves of the same are equal to one another.
8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
9. The whole is greater than its part.
10. Two straight lines cannot enclose a space.
11. All right angles are equal to one another.
12. If a straight line meets 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.
PROBLEM. 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.
From the centre A, at the distance AB, describe (3 Postulate) the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines (1 Post.) CA, CB, to the points A, B: ABC shall be an equilateral triangle.
Because the point A is the centre of the circle BCD, AC is equal (15 Definition) to AB; and because the point B is the centre of the circle ACE, BC is equal to BA:
But it has been proved that CA is 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 (1 Axiom); therefore CA is equal to CB; wherefore CA, AB, BC 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.
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.
From the point A to B draw (1 Post.) the straight line AB; and upon it describe (1. 1.) the equilateral triangle DAB, and produce (2 Post.) the straight lines DA, DB, to E and F; from the centre B, at the distance BC describe (3 Post.) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL: AL shall be equal to BC.
Because the point B is the centre of the circle CGH, BC is equal (15 Def.) to BG;
And because D is the centre of the circle GKL, DL is equal to DG, and (Constr.) DA, DB, parts of them, are equal; therefore the remainder AL is equal to the remainder (3 Ax.) 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 (1 Ax.) 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.
PROB. 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.
From the point A draw (2. 1.) the straight line AD equal to C; and from the centre A, and at the distance AD describe (3 Post.) the circle DEF;
And because A is the centre of the circle DEF, AE is equal (15 Def.) to AD; but the straight line C is likewise equal (Constr.) to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal (1 Ax.) to 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.
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 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 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 angle BAC equal to the angle EDF, 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 angle DEF, and the angle ACB to DFE.
For, if the triangle ABC be applied to DEF, so that the point 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 (Hyp.) to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal (Hyp.) to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC (Hyp.) 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 enclose a space, which is impossible (10 Ax.). Therefore 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 (8 Ax.); 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.
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 equal to AC, and let the straight lines AB, AC be produced to D 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 AG equal to (3. 1.) AF, the less, and join FC, GB.
Because AF is equal (Constr.) to AG, and AB to (Hyp.) 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; therefore the base FC is equal (4. 1.) to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal (4. 1.) 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:
And because the whole AF is equal to the whole AG (Constr.), of which the parts AB, AC, are equal (Hyp.); the remainder BF shall be equal (3 Ax.) 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;