DEFINITION. A circle is said to be inscribed in a rectilineal figure when the circumference of the circle touches each side of the figure. Bisects PQR, PRQ by QC, RC, intersecting in C. shall be the the described with centre Cat the distance CD required. QDC, DQC, and the side QC are respec tively = LS QFC, FQC, and the side QC, :. DC = FC. Similarly it may be proved that DC = EC. (Ι. 25) In the given draw two diameters PCQ, RCS at right 2s to one another. Join PS, SQ, QR, RP. Then PSQR shall be the square required. *. the 2 s at Care right s; ... they are = one another; and... the chords PS, SQ, QR, RP on which they stand are equal; ... PSQR is equilateral. (111. 18) Again, ... RPS is in a semicircle; ... it is a right 2. (III. 16) Similarly the other 2 s of PSQR are right ... PSQR is rectangular, and it has been proved to be equilateral; ... PSQR is a square and it is inscribed in the given. In the given draw two diameters PCQ, RCS at right 2 s to one another: at P, S, Q, R draw tangents intersecting in V, X, Y, Z. VXYZ shall be the square required. the sat Pand Care right s; (III. 6) ... VPX is || to RCS. (I. 20) Similarly ZQY is || to RCS; (Ι. 22) ... VPX is || to ZQY. So VRZ, XSY are each || to PCQ and are ... || to one another. Then VX and ZY are each = RS; ·· VS, SZ are also VZ, XY are each = PQ ; ·.· VQ, PY are but RS = PQ; .·. VX, ZY, VZ, XY are ... VY is equilateral. = S: S: one another; (Ι. 26) Also angle at X = L PCS ; ·.· PXSC is a .. at X is a right ; so sat Y, Z, Vare right s; ... VV is rectangular. Hence VV is a square and it is described about the Draw the diagonals AC, BD intersecting in Q. · AB = BC ; ..ᄂBAC = L BCA ; (1. 2) and since ABC is a right ; .. LS BAC, BCA are each of them = half a right 2. (1. 24) Similarly ABD = half a right 4, and .. = 2 BAC; ... AQ = BQ. Similarly BQ = CQ = DQ; ... the passes through B, C, D ; and is... described about the square ABCD. (Ι. 4) Let ABCD be the given square: it is required to inscribe a circle in ABCD. Bisect AB, BC in P, Q. Through P, Q draw PR, QS || to BC, AB, respectively. With O as centre and distance OP describe a: it shall be the one required. *** the opposite sides of s are = one another; (1. 26) •*. SO = AP, OQ = PB, PO = QB, and OR = QC; . SO, OQ, PO, OR are all = one another; ... the will pass through Q, R, S: also. 4 SAP is a right.. OPA is a right 2. (1. 21) Similarly the sat Q, R, S are right ∠s; ... the sides of the square touch the. (111. 6) We now proceed to solve the general problems of inscribed and circumscribed regular figures, the constructions in the case of squares having been given separately on account of their simplicity. |