Then, fince an angle at the centre of a circle is double to that at the circumference (III. 14.), the angle BAC will be half the angle bec. And, for the same reason, the angle BDC will, also, be half the angle BEC. But things which are halves of the same thing are equal to each other ; consequently the angle bac is equal to the angle BDC. Again, let the segment BADC be not greater than a semicircle : Then, since the right lines BD, ac interfect each other in F, the angle BFA will be equal to the opposite angle DFC (I.15.) And because the fegment ABCD is greater than a femi. circle, the angles ABD, ACD, which stand in that lege ment, are equal to each other (III. 15.) But since the two angles BfA, ABF of the trianglo FBA, are equal to the two angles DFC, FCD of the triangle DCF, the remaining angle BaF, or BAC, will also be equal to the remaining angle. FDC, or BDC. Q. E. D. An angle in a semicircle is a right angle, FA B Let ABC be a semicircle ; then will any angle ACB, in that semicircle, be a right angle. For, find the centre of the circle e (III. 1.) and draw the diameter ced. Then, because an angle at the centre of a circle is double to that at the circumference (III. 14), the angle AED will be double the angle ACD. And, for the same reason, the angle bed will be double the angle BCD. The angles AED, BED, therefore, taken together, are double the whole angle ACB. But the angles AED, BED, are, together, equal to two. right angles (1.13.); consequently the angle ACB will be equal to one right angle. Q. E.D. COROLL. The angle BAC, which stands in a segment greater than a semi-circle, is less than a right angle (I. 28.): And the angle BCF, which stands in a segment less than semi-circle, is greater than a right angle. PRO P. XVII. THEOREM. The opposite angles of any quadrilateral figure, inscribed in a circle, are equal to two sight angles. Let ABCD be a quadrilateral, inscribed in the circle ADCB; then will the opposite angles BAD, BCD, taken together, be equal to two right angles. For, For, draw the diagonals ac, BD, and produce the fide BA to E. Then, because the outward angle of any triangle, is equal to the two inward opposite angles taken together (I. 28.), the angle EAD will be equal to the angles ABD, ADB. And, because all angles in the same segment of a circle are equal to each other (III. 15.), the angle ABD will be equal to the angle ACD, and the angle ADB to the angle ACB. The angle EAD, therefore, which is equal to the angles ABD, ADB, taken together, will also be equal to the angles ACD, ACB, taken together, or to the whole ana gle BCD. But the angles EAD, BAD, taken together, are equal to two right angles (I. 13.); consequently the angles BCD, BAD, taken together, will also be equal to two right angles. Q. E. D. COROLL. If any side AB, of the quadrilateral ABCD, be produced, the outward angle EAD will be equal to the inward oppofite angle BCD. Through any three points, not situated in the same right line, to describe the circumference of a circle. B D A H Let A, B, C, be any three points, not situated in the same right line ; it is required to describe the circuma. ference of a circle through those points. Draw the right lines AB, BC, and bisect them with the perpendiculars, DH, EG (I. 10 and 11.); and join DE. Then, because the angles FED, FDE are less than two right angles, the lines DH, EG will meet each other, in some point F (Cor. I. 25-); and that point will be the centre of the circle required. For, draw the lines Fa, FB and Fc. Then, fince Ad is equal to DB, DF common, and the angle ADF equal to the angle FDB (I. 8.), the fide FA will also be equal to the side FB (I. 4.) And, in the fame manner, it may be thewn, that the fide Fc is also equal to the side FB. The lines FA, FB and Fc, are, therefore, all equal to each other; and consequently F is the centre of a circle which will pass through the points A, B and c, as was to be shewn. 4 Scho. Scho. If the segment of a circle be given, and any three points be taken in the circumference, the centre of the circle may be found, as above, PRO P. XIX. THEOREM. If the opposite angles of a quadrilateral, taken together, be equal to two right angles, a circle may be described about that quadria lateral. Let ABCD be a quadrilateral, whose opposite angles DCB, DAB are, together, equal to two right angles : then may a circle be described about that quadrilateral. For fince the circumference of a circle may be defcribed through any three points (III. 18.), let e be the centre of a circle which passes through the points D, c B; and draw the indefinite right line era. And if the circle does not pass through the fourth point A, let it pass, if possible, through some other point F, in the line EA, and draw the lines DF, FB, and BD. Then, since the opposite angles BFD, DCB are, together, equal to two right angles (III. 17.), and the angles BAD, DCB are also equal to two right angles (by Hyp.), the angles BFD, DCB will be equal to the angles BAD, DCB. |