For, that the triangle KLM is circumscribed about the circle is plain from the construction, because each of its sides is made to touch the circle. It is also similar to DEF; for, in the quadrilateral AOBL the angles at A and B are both right angles, but all the angles of a quadrilateral figure are together equal to four right angles,.. AOB+ ▲ ALB = two right angles = - ᎠᎬᏳ + ▲ DEF (30). But AOB ▲ DEG, .. L ALB, or ▲ KLM = 2 DEF. Similarly KML = DFE, and .. 4 LKM = 2 EDF (37), that is, the triangle KLM is equiangular to the triangle DEF. = 153. PROP. LV. In a given triangle to inscribe a circle. Let ABC be the given triangle; see fig. (140.) Bisect the angles BAC, ABC by the straight lines AD, BD, intersecting in D. From D draw DE perpendicular to AB. With centre D, and radius DE, describe a circle, and it shall be the circle required. For, drawing DF perpendicular to BC, and DG perpendicular to AC, it is easily shewn that DE=DF=DG; .. the circle described passes through E, F, and G. And the angles at those points are right angles, .. AB, BC, and AC touch the circle. This construction contains the solution of the following Problem: : To cut the greatest circle out of a given triangle'. 154. PROP. LVI. About a given triangle to circumscribe a circle. This is the same construction as that which occurs in (134) where we construct a circle to pass through three given points. Here the given points are the vertices of the three angles of the given triangle. 155. PROP. LVII. In a given circle to inscribe a square. Let ABCD be the given circle; O the centre; draw two diameters AC, BD at right angles to each other. Join AB, AD, CB, ČD, and ABCD is the square required. For, since AC, BD are diameters B of the circle, each of the angles of the figure ABCD is the 'angle in a semicircle', and is .. a right angle (54). Also AC, BD divide the circumference C into four equal parts, .. AB, AD, CB, CD are chords of equal arcs, and .. are equal to one another. COR. If each of the arcs AB, BC, CD, DA be bisected (133), and chords be drawn from the corners of the square to each point of division, a regular octagon will be inscribed in the circle. This construction shews us how to cut the greatest square, or octagon, out of a given circle'. 156. PROP. LVIII. About a given circle to circumscribe a square. F E Let ABCD be the given circle; O the centre; draw two diameters AC, BD at right angles to each other; through the points A, B, C, D draw FE, FG, GH, HE touching the circle; EFGH is the square required. For, by (55) the angles at A, B, C, D are right angles; also, by construction the angles at O are right B C H angles; and all the angles of any four-sided figure are together equal to four right angles: .. each of the angles at E, F, G, H is a right angle. Again, by (34 Cor. 2) FE is parallel to BD, and to GH; also FG is parallel to AC, and to EH ; .. EFGH is a parallelogram, .. EF= GH BD (40); and FG-EH-AC; but AC-BD, :. EF = FG = GH = HE; and .. EFGH is a square. = 157. PROP. LIX. In a given square to inscribe a circle. See fig. in last Prop. Let EFGH be the given square. Bisect each of the sides EF, FG in A and B; through A draw AC parallel to FG; and through B draw BD parallel to EF, intersecting AC in O; then with centre O, and radius OA, describe a circle, and it shall be the circle required. For, since each of the foursided figures are, by construction, parallelograms, it may easily be shewn that OA = OB=OC = OD, and that the angles at A, B, C, D are right angles. Then it follows, that the circle touches each of the sides of EFGH, and is .. 'inscribed' in it. This construction shews us how to cut the greatest circle out of a square'. 158. PROP. LX. About a given square to circumscribe a circle. See fig. in (155). Let ABCD be the given square. Draw the diagonals AC, BD, intersecting in O. With centre O, and radius OA describe a circle, and it shall be the circle required. For, by (40 Cor. 2) AC, BD bisect each other in 0; .. since ACBD, OA = OC = OB=OD, and ... the circle with centre O, and radius OA, will pass through B, C, D, as well as A. 159. PROP. LXI. In a given circle to inscribe a regular pentagon. Let O be the centre of the given circle; draw OA any radius, and OB another radius at right angles to OA, so that AB is an arc of a quadrant of the given circle. Divide the arc AB into D five equal parts; and let AC be the first of such parts, reckoned from A. Join BC; then draw the chords BD, CF, FE, each equal to BC; and join ED. BDEFC shall be the pentagon required. For, by construction it is a five-sided figure; and four of the sides are made equal to one another. Also, ED, the remaining side, is the chord of the arc, which remains after taking from the whole circumference four-fifths of This may be done by trial with the compasses; see (169). = the arc of a quadrant four times; for BC is four-fifths of AB, by construction; that is, the arc ED-the difference between four quadrants and 3 quadrants four-fifths of a quadrant-arc BC; . BDEFC is equilateral. It is also equiangular, since each angle is the angle in one of five segments all of which are equal to one another. The accuracy of the work may be tested by observing whether the diagonals DF, DC, BE, BF are all equal to one another, as they ought to be. COR. If G be the third point of division from A, that is, the second from B, by joining BG and continuing equal chords round the circle, we inscribe a regular decagon in the circle. This construction shews us how to cut the greatest regular pentagon or decagon out of a given circle'. 160. PROP. LXII. In a given circle to inscribe a regular hexagon. Let O be the centre of the given circle; draw OA any radius; and beginning from A, draw AB, BC, CD, DE, EF five consecutive chords each equal to OA; and join FA. ABCDEF is the hexagon required. For, by construction it is a sixsided figure, and five of its sides are made equal to one another. Also, joining OB, OC, OD, OE, OF, it is plain that each of the triangles E A OAB, OBC, OCD, ODE, OEF is equilateral, and .. each of the five angles at O is one-third of two right angles (37), that is, one-sixth of four right angles; and .. the sum of them is five-sixths of four right angles. But < AOF makes up the whole four right angles about the common vertex O (30 Cor. 2), .. ▲ AOF = one-sixth of four right angles; and .. the chord AF-each of the other chords; and ABCDEF is equilateral. Again, it is equiangular, because all the angles are angles in equal segments of the same circle. COR. If BD, BF, DF be joined, BDF is an equilateral triangle inscribed' in the circle. 6 This construction shews us how to cut the greatest regular hexagon, or equilateral triangle, out of a given circle'. 161. PROP. LXIII. In a given regular polygon of any number of sides to inscribe a circle. Let AB be a side of the given polygon; and bisect each of its angles at A and B by the straight lines AO, BO, intersecting in O. From O draw OC, perpendicular to AB; with centre O, and radius OC, describe a circle, and it shall be the circle required. For the proof of this, see (87 Cor. 3). 162. PROP. LXIV. About a given regular polygon of any number of sides to circumscribe a circle. Let AB be a side of the given polygon, as in (161); and bisect each of its angles at A and B by the straight lines AO, BO, intersecting in O. With centre O, and radius OA describe a circle, and it shall be the circle required. For the proof of this, see (87 Cor. 1). 163. PROP. LXV. To construct a regular pentagon with each of its sides equal to a given line or length. Let AB be the given straight line or length; with centres A and B, and radius AB, describe two arcs, somewhat greater than those of quadrants, on that side of AB on which the pentagon is to lie, and two small arcs on the other side intersecting in the point I. Draw the indefinite line ID, through the intersections of these arcs; and through B draw BH parallel to ID, and meeting the circumference of one of the circles in H, so that ABH is a quadrant. Divide the arc AH into five equal parts, marking the B second division F, and the fourth G, reckoning from A. Make the arc HC, on the other side of H, equal to the arc HG (132). Join BC, BG, BF; produce BF to meet |