VII. The whole is greater than its part. VIII. The parts together are equal to the whole. IX. Two straight lines can not enclose a space. X. If two straight lines intersect one another they cannot be both parallel to the same straight line. XI. All right angles are equal. XII. If one straight line be placed on another they wholly coincide, that is, become one straight line. PROP. I. THEOREM. *If a straight line passes through the centre of a circle and is terminated both ways by the circumference, the straight line divides the circle into two equal parts. Let ABCD be a circle and F its centre. Let AC a straight line pass through the centre and be terminated both ways by the circumference at the points A and C. AC divides the circle ABCD into two equal parts ABC and ADC. A * Given a circle, the centre and a diameter, to prove that the diameter bisects the circle. This is simply the converse of the definition of a circle. The two corollaries are also to be proved by the same definition. For if ADC be applied to ABC, they coincide. But things which coincide, that is, exactly fill the same space are equal (Ax. VI). But if ADC do not coincide with ABC, let them stand as indicated. Join FD, and produce FD to B. Because a circle is a plane figure bounded by one line and is such that all straight lines drawn from the centre to the circumference are equal (Def. 6), therefore all the straight lines drawn from F to the different points in the circumference ABCD are equal; therefore FD, which is drawn from F the centre to D a point in the circumference, is equal to FB which is drawn from F the centre to B a point in the circumference. But FD is a part of FB. And the part is equal to the whole, which is impossible (Ax. 7). Therefore ADC will coincide with ABC. Wherefore, if a straight line passes through the centre of a circle, &c. Q.E.D. Corollary.-If one circle touch another internally, they have not the same centre (Prop. 5: 3. E.). Cor. 2.—If two circles cut one another, they have not the same centre (Prop. 6 : 3. E.). PROP. II. (Prop. 15. Book I. E.) If two straight lines cut one another, the vertical or opposite angles are equal. Let the straight lines AC and BD cut one another at the point F. The vertical or opposite angles AFB and CFD are equal to one another, and so are the angles BFC and AFD. From the centre F with any of the straight lines AF, BF, FC, FD draw the circle ABCD and complete the diameters AC and BD. Because AC is a diameter, therefore it divides the circle ABCD into two equal parts ABC and ADC (P. 1); and ABC is a half of the circle. Again because BD is a diameter therefore it divides the circle into two equal parts BAD and BCD (P. 1); and BCD is a half of the circle. But equal parts of the same thing are equal (Ax. 5.) Therefore ABC and BCD are equal. Take away the common part BFC and the remainders AFB and CFD are equal (Ax. 3). Apply AFB to CFD, so that the point F of the radius FB may be on the point F of the radius FC and the point B on the point C. Therefore the radii FB and FC coincide (Ax. 12). And FB coinciding with FC, the radius FA coincides with the radius FD and the arc BA with the arc CD. Therefore the sector AFB coincides with the sector CFD. 1 For, if, AFB do not coincide with CFD, first, let a part of the radius FA fall beyond the arc CD. But that is impossible. Because, in that case, a part of the radius FA would be equal to the whole radius, as has been shown in the first proposition. Secondly, let the radius FA remain within the arc CD but the arc BA (i.e. CGA) fall beyond the arc CD. Join FG. In this case also, FH which is a part of the radius FG is equal to the whole radius FG. This is therefore impossible. Lastly, let the radius FA remain within the arc CD and the arc BA coincide with a part of the arc CD. This is also impossible. Because the sector CFD fills more space than the sector AFB, and therefore they are unequal (Ax. 6), while it has been shown that they are equal. Therefore the sector AFB coincides with the sector CFD and the angle AFB and CFD coincide and are therefore equal (Ax. 6). In this way, it can be shown that the angles BFC and AFD are equal. Wherefore, if two angles, &c. Q.E.D. PROP. III. THEOREM. (Prop 4. Book I. E.) 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 these sides equal to one another, their bases or third sides are equal the two triangles are equal and their other angles are equal, each to each; viz. those to which the opposite sides are equal. : Let ABC, DEF be two trianglès, which have the two A A B base BC is equal to the sides AB, AC equal to the two sides DE, DF, each to For, if the triangle ABC be applied to the triangle straight line AB upon the straight line 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. Also, the point C shall coincide with the point F, because AB is equal to DF. But the point B was proved to coincide with the point E. Therefore the base BC shall coincide with the base EF. For, the point B coinciditng with the point E, and the point C with the point F, if the base BC does not coincide with the base EF, the two straight lines BC, EF would enclose a space, which is impossible (Ax. 9) wherefore the base BC coincides with the base EF and is, therefore, equal to it. Wherefore, also the whole triangle ABC coincides with the whole triangle DEF, and is, therefore equal to it. angles of the one coincide with the the other and are therefore equal to And the remaining remaining angles of them, each to each; viz., the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Wherefore if two triangles have two sides of the one &c. Q. E. D. PROP. IV. THEOREM. (Prop. 5 and 6: 1. E.) The angles at the base of an isosceles triangle are equal to one another; and conversely, if the angles at the base of a triangle are equal, it is an isosceles triangle. First, let ABC be an isosceles triangle, having the sides AB and AC equal. The angle at B is equal to the angle at C. D From the centre A with AB or AC, draw a circle BCFD. Produce BA and g CA and make AF and AD equal to BA or CA (def. 6.) Join DF. |