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BCE is double of BDC; and consequently the angle A CB is double the angle ADB. Q. E. D.
Cor. 1. As the angle A C B, at the centre, is measured by the arc A B on which it stands, the angle A DB, at the circumference, is measured by half the arc AB on which it stands.
Cor. 2. All angles in the same segment of a circle, or standing on the same arc, are equal to each other ; for each of them is equal to half the angle at the centre, standing on the same arc.
Cor, 3. The angle in a semicircle is a right angle. For when AC and C B (first fig. of this prop.) become one straight line, A D B will be a semicircle, and the angles ACE and BC E together, will be equal to two right angles ; whence AD B, which is half the sum of ACE and BCE, will in that case be equal to one right angle.
THEOREM LIV. The sum of any two opposite angles, B AC, and BCD, of a quadrangle ABCD inscribed in a circle, is equal to two right angles.
For the angle BAC is measured by half the arc BCD, and the angle BCD by half the arc BAD; (Cor. 1. Theo. 53.) therefore the sum of the angles BAC and BCD, is measured by half the whole circumference. But half the whole circumference
E A is the measure of two right angles. Hence the angles B A C and BC D are, together, equal to two right angles. Q. E. D.
THEOREM LV. If any side B A, of a quadrilateral, ABCD, (see the last figure) inscribed in a circle, be produced, the outward angle, DAE, will be equal to the inward and opposite angle, C.
For the angles DAB and D A E are equal to two right angles, (Theo. 6.) and the angles D A B and DCB are also equal to two right angles. (Theo. 54.) Hence by omitting the common angle DAB from each sum, the remaining angles, D AE and DC B, are equal. Q. E.D.
THEOREM LVI. If, in a circle ABD C, there be drawn two parallel chords A B, CD; the intercepted arcs A C and B D will be equal. For join BC. Then because A B and C D are
G F parallel, the alternate angles A B C and DCB are
B equal. (Theo. 17.) But the angle A B C is mea
D sured by half the arc A C, and the angle DCB, by half the arc D B. (Cor. 1. Theo. 53.) Whence the halves of AC and B D are equal, and consequently the whole arcs A C and B D are equal. Q. E. D.
Cor. If A B move parallel to itself, till it coincide, at G, with the tangent EGF, the intercepted arcs, G C and G D, will be equal.
If A B be a tangent to a circle, and AC a chord drawn from A, the point of contact, the angle BAC will be equal to any angle in the alternate segment ADC. For from C draw the chord D C parallel to the
B tangent A B, and join A D. Then as the arcs AC and AD are equal, (Theo. 56.) the angles ADC and AC D are equal, being measured by the halves
DI of the equal arcs AC and A D. (Cor. 1. Theo. 53.) But the angle ACD is equal to the alternate angle BAC (Theo. 17.) ; therefore the angle BAC is equal to the angle AD C. And as the angle ADC is equal to any angle in the segment ADC (Theo. 53. Cor. 2.), the angle B A C is also equal to any angle in the alternate segment ADC. Q. E.D.
Cor. As the angle ADC is measured by half the arc A C, the angle B AC, made by the tangent A B, and the chord A C, is also measured by half the intercepted arc A C.
The angle D E B formed within any circle ABC D, by the intersection of two chords A B, CD, is measured by half the sum of the intercepted arcs A C and BD. For join A D. Then the angle ADE or ADC is
C measured by half the arc A C, and the angle D A E, or
D measured by half the sum of the arcs A C and B D.
B Q. E. D.
If two chords A B, C D, of the circle A B C D, meet when produced in a point E without the circle, the angle E is measured by half the difference of the intercepted arcs A C and B D.
For join AD. Then the angle ADC is measured by half the arc A C, and the angle D A B by half the arc D B. (Cor. 1. Theo. 53.) But the angle E is equal to the difference of the angles DAB and EDA (Theo. 22. Cor.); and therefore the angle D E B is measured
Cor. If one of the lines, as ECD, revolve round the point E, till the points C and D coincide in F, EF will then be a tangent to the circle, and the angle FE A will be measured by half the difference of the intercepted arcs A F and F B.
If A B, any chord of a circle, be bisected in D, and the point D be joined to C, the centre of the circle ; then C D will be perpendicular to A B; or if CD, drawn from the centre, be perpendicular to the chord A B, AB will be bisected in D.
For let the two radii CA, CB be drawn, then if A B be bisected in D, the two triangles A CD, BCD, will have the sides A C and B C equal, A D and BD
C equal, and D C common to both ; they are therefore
D B identical (Theo. 5.), and have the adjacent angles ADC and BDC equal; these angles are conse
E quently right angles, and C D is therefore perpendicular to A B.
Again if C D be perpendicular to AB, the angles CD A and CDB will be equal ; and C A being equal to C B, the angle C AB, or CAD is equal to the angle C BA, or CBD (Theo. 3.); therefore ACD and BCD, the remaining angles of each triangle, will also be equal (Theo. 24. Cor. 1.) Hence, as AC is equal to C B, the triangles ADC and BDC will be identical (Theo. 2.), and consequently the side A D will be equal to the corresponding side B D; or A B is bisected in D. Q. E. D.
Cor. Since the angle ACD is equal to the angle BCD, the arc A E will be equal to the arc B E.
Let A B, C D be any two chords in a circle A B D C, and let G E, GF be perpendiculars drawn from the centre G, on the chords A B, and C D. If A B and C D are equal, E G and G F are equal; or if E G and GF are equal, A B and C D are also equal.
For draw the two radii A G, GC; then as A B and А с CD are bisected in E and F (Theo. 60.), if they are equal, their halves A E and C B will be equal. But
F the square of AG is equal to the sum of the squares
G of A E and EG, and the square of G C is equal to the sum of the squares of G F and FG (Theo. 44.) ; and B В as A G and G C are equal, their squares are equal. Hence the squares of A E and E G are together equal to the sum of the squares of CF and FG; and if the squares of the equal lines A E and C F be taken from each sum, the square of E G will remain equal to the square of GF, and consequently GE will be equal to GF. In the same way it may be shewn, that when GE is equal to GF, AE is equal to CF, therefore A B and C D, the doubles of those lines, are equal.
If B D and B C be two unequal chords in a circle, the angle B A D at the centre, subtended by the greater chord BD is greater than the angle BAC subtended by the less.
For as the sides A D, A B, of the triangle D AB are equal to the sides A C, AB, of the triangle
A CAB, but D B is greater than BC, therefore the angle DAB is greater than the angle CAB. D
B В (Theo. 14.)
Cor. 1. As the angle D A B is measured by the arc DC B, and the angle CAB by the arc BC, therefore the arc DCB, subtended by the greater chord, is greater than the arc B C, subtended by the less.
Cor. 2. In the same circle equal chords subtend equal arcs, or equal angles, whether at the centre or circumference; and equal arcs, or equal angles, whether at the centre or the circumference, are subtended by equal chords.
If ACB, an angle at the centre of a circle, be the sixth part of four right angles, or the third part of two right angles, A B, the chord of the arc which measures the angle ACB, will be equal to the radius of the circle.
For as all the interior angles of a triangle are, together, equal to two right angles, (Theo. 24.) if ACB
C be the third part of two right angles, the remaining angles A and B will, together, be equal to two-thirds of two right angles. And as A C and B C are equal,
B the angles A and B are equal (Theo. 3.); and consequently each of these is equal to the third part of two right angles, or to the angle ACB. Hence the triangle ABC being equiangular, is equilateral, (Cor. Theo. 4.) or A B is equal to AC or BC.
Remark. The circumference of a circle, or the measure of four right angles, is commonly divided, in practice, into 360 equal parts, called degrees ; therefore the chord of 60 degrees, or the chord of the measure of the sixth part of four right angles, is in any circle, equal to the radius ; and hence if the chords of every arch in a circle of a given radius, were arranged on a line, such a line, called a line of chords, would afford a convenient practical method of describing THEOREM LXIV. If A B. be the radius of a circle, then B C, a perpendicular on its extremity B, will be a tangent to the circle. For from any other point C, in the line B C, draw
B с CDA to the centre A, meeting the circle in D. Then because the angle A B C is a right angle, the angle ACB is less than a right angle (Theo. 24.) ; therefore A AC is greater than A B (Theo. 12.), or than its equal AD; and consequently the point C is without the circle. In the same way every other point of the line BC, except the point B, may be shewn to be without the circle ; and therefore the line B C meets the circle in the point B only, and it is consequently a tangent to the circle. Q. E. D.
THEOREM LXV. If B C (see the last figure) touch the circumference of the circle in B, the radius A B will be perpendicular to BC.
For as BC touches the circumference in the point B only, every other point of B C will be without the circle; A B is therefore the shortest line that can be drawn from the point A to the line BC; hence A B is perpendicular to B C. (Theo. 27.)
THEOREM LXVI. If B D touch the circumference of a circle in B, then B A drawn perpendicular to BC, will pass through the centre of the circle.
For if the centre of the circle be not in the line A B, let any point C out of that line be the centre of
A C the circle, and join BC. Then the angle C B D is a right angle (Theo. 64.), and it is consequently equal to ABD, which is also, by condition, a right angle;
B D that is, the less, angle is equal to the greater, which is impossible. Hence, no point out of the line B A can be the centre of the circle, and the line B A therefore passes through the centre. Q. E. D.
THEOREM LXVII. In
any circle ACDB, if the chord CD, and the diameter A B, meet each other in G, the rectangle of GA, G B, the segments of the one, will be equal to the rectangle GC, G D, the segments of the other. For let D E be joined, and the
G perpendicular E F drawn to DC. Then because DE is equal to E B, G B is the sum of G E and ED; and because D E is equal to E A, AG
F is the difference of D E and E G.
E Again, because D F is equal to FC D
DS (Theo. 60.), G C is the difference of
B the segments of the base D F and FG, made by the perpendicular EF; and D G is the sum of the segments D F and F G.