Because the S QDC, DQC are equal to the s QFC, FQC, .. LQCD = 4 QCF; also QC is common to the As DQC, FQC, .. CD= CF Similarly CD CE. .. the circle passes through Fand E. (I. 20) (I. 6) Again, because QR is perpendicular to CD, (III. 3) .. QR touches the circle. Similarly RP and PQ touch the circle. Wherefore a circle has been inscribed in the given triangle. Q.E.F. DEFINITION. A rectilineal figure is said to be inscribed in a circle when all its angular points are upon the circumference of the circle. PROPOSITION III. To inscribe a square in a given circle. P R Q Let PRQ be the given circle; it is required to inscribe a square in it. Draw the two diameters PCQ, RCS perpendicular to one another. Join PS, SQ, QR, RP. Then PRQS shall be the square required. Because PC, CR, and PCR are respectively equal to QC, CR, and QCR, ... PR=RQ. Similarly RQ QS, and QS= SP. .. PRQS is equilateral. (I. 4) Again, since RPS is in a semicircle, .. it is a right. angle. ... (III. 6) Similarly the others of PRQS are right angles. .. PRQS is a square. Wherefore a square has been inscribed in the given circle. Q.E.F. DEFINITION. A regular hexagon is a plane rectilineal figure bounded by six equal straight lines, and having its angles equal to one another. PROPOSITION IV. To inscribe a regular hexagon in a given circle. T U X Let PQR be the given circle; it is required to inscribe a regular hexagon in it. Let the circumference be divided into six equal arcs PXQ, QYR, RZS, &c. (III. II, Cor.) Draw the chords PQ, QR, RS, &c. For if the figure were turned about C, so that P might coincide with the former position of Q, and.. Q with that of R, R with that of S, and so on, since the arcs PXQ, QYR, &c., are all equal, then the chord PQ would coincide with QR, and is ... = it : = similarly QR QS, and so on. .. the hexagon is equilateral. Also the arms of the of QRS were. PQR would fall where those Hence PQR = LQRS. Similarly ▲ QRS = ▲ RST; and so on. ..the hexagon is equiangular. Wherefore a regular hexagon &c. Q.E.F. |