2 P. In a given circle (ABC), to inscribe a triangle equiangular to a given triangle (DEF). Construction. Draw the tangent GAH (a); at A, the point of contact, make the angle HAC equal to the angle E, also the angle GAB equal to the angle F (b); join BC. G A H Argument. Because the straight line GAH touches the circle ABC, and from the point of contact A, the straight lines AB, AC are drawn, the angle HAC is equal to B, and the angle GAB is equal to C, in the alternate segments of the circle (c). But HAC is made equal to E, and GAB to F, of the given triangle; therefore, the angle B is equal to E, and the angle C is equal to F, (d); so the third angles BAC and D must be equal (e). Therefore, the triangle ABC has all its angles equal to those of the given triangle DEF, each to each; and it is inscribed in the given circle ABC, which was to be done. Recite (a) p. 17, 3; (d) ax. 1; (b) 23, 1; (c) p. 32, 3; 3 P. About a given circle (ABC), to describe a triangle equiangular to a given triangle (DEF). Produce EF both ways to the points G, H: find K, the centre of the given circle (a); draw any radius KB; at the point K, in BK, make angles BKA, BKC equal to the exterior angles DEG, DFH, each to each (b); draw tangents through the points A, B, с (c), to meet in the points M, L, N. There fore LM, LN, MN, meet the radii at right MBN angles in the points A, B, C (d): and because D L GE FH AKC the quadrilateral AMBK may be divided into two right angled triangles, whose angles are equal to four right angles (e); and that two of its angles at A and B, are right angles, the other two, AKB, AMB, are equal to two right angles; but the two angles DEF, DEG are also equal to two right angles (f), and BKA was made equal to DEG; therefore, the remaining angles DEF, AMB are equal. It may be proved, in this way, that the angles DFE and N are also equal; and so the third angles Dand L must be equal (e). Wherefore, about given circle a triangle is described equiangular to a given triangle; which was to be done. a Recite (a) p. 1, 3; (d) p. 18, 3; (b) p. 23, 1, (c) p. 17, 3; 4 P. To inscribe a circle in a given triangle (ABC). Constr. Bisect the angles B and C (a), by straight lines BD, CD, meeting in D, from which point draw perpendiculars to meet the sides of the triangle in the points E, F, G (b). A Argument. The triangles BDE, BDF are equal, E for the following reasons, viz. The angles EBD, FBD, are halves of the angle EBF (c); BED, BFD are right angles (d), and BD is common to the two triangles: therefore DE is equal to DF B (e): for like reason, DG is also equal to DF, or DE; and the circle described upon the centre D, at the distance of any of them, will pass through the points E, F, G; and be inscribed in the given triangle (f); which was to be done. Recite (a) p. 9, 1; (b) p. 12, 1; (d) ax. 10; (e) p. 26, 1; (c) ax. 7; 5 P. To describe a circle about a given triangle (ABC). Constr. In points D, E, bisect the sides AB, AC, of the given triangle (a); draw DF, EF, at right angles to the sides (b); and, if the point F be c on the side BC, join FA; if within or without the triangle, join also FB, FC. Argument. Now, on account of the right angles at D (c), and that DA, DF are equal to DB, DF, the bases AF and BF are equal (d). In like manner it may be proved that FC is equal to FA, or FB: therefore F is the centre of a circle passing through the three points of the triangle, as required (e). Recite (a) p. 10, 1; (c) ax. 10, 1; B (d) p. 4, 1; Corollary. The point F is within, on the side, or without the triangle, according as it may be acute, right, or obtuse angled. 6 P. To inscribe a square in a given circle (ABCD). Draw, in the circle, the diameters AC, BD, at right angles to each other (a); they will divide the circumference into four equal arcs (b); draw also the chords AB, AD, CB, CD, which are all equal to each other (c). Now each of the angles made by the chords, at A, B, C, D, is in a semicircle, and is therefore a right angle (d). Wherefore, the quadrilateral ABCD, has four equal sides and four right angles, and is there- B fore a square (e), inscribed in the given circle. Recite (a) def. 10, 1; (b) p. 26, 3; (c) p. 29, 3; (d) p. 31, 3; (e) def. 30, 1. 7 P. To describe a square about a given circle (ABCD). B A F E D Draw the diameters AC, BD, at right angles C to each other (a); also tangents (b) to the circle, through the points A, B, C, D, at right angles to the diameters, meeting in the points F, G, H, K. Now, because of the equal alternate angles (c) GBD, BDK, the side GH is parallel to FK: for the same reason GF is parallel to HK: therefore FGHK is a parallelogram, whose opposite sides and angles are equal (d): but because any one of the angles at E and an interior angle on the same side of those at the points A, B, C, D, are equal to two right angles (e); therefore each of the angles F, G, H, K, is a right angle. Wherefore, since the equal sides touch the circle and meet at right angles, there H C K is a square described about the circle as was required (f). Recite (a) cor. p. 1, 3; (b) def. 2, 3; (c) p. 27, 1; (d) p. 34, 1; (f) def. 30, 1, and 2, 4. 8 P. To inscribe a circle in a given square (ABCD). A Bisect two sides AB, AD, of the given square, in the points E, F (a); draw EH, FK parallel to AB, AD (6): then the interior angles A, E and also A, F, are equal to two right angles (c): for E same reason all the angles about the point G are right angles; therefore the opposite sides GE, GF, GH, GK are equal (d), and G is the centre of a circle that shall touch the sides of the given. B the square, in E, F, H, K (e), and be inscribed therein, as required (f). 9 P. To describe a circle about a given square (ABCD). Join the opposite angles of the given square by the diameters AC, BD, intersecting each other in Es Then the triangles ABC, ADC, have two sides AB, AD, equal, and AC common; also the bases BC, DC are equal; therefore the angle CAB equals the angles CAD (a). C In like manner, it is shown, that the other angles of the square are bisected by the straight lines B AC, BD, (b): but the angles of the square are all equal to each other (c); therefore the halves BDA, BDC, ACB, ACD, DBA, DBC are also equal to each other (d); and their bases EA, EB, EC, ED are equal (e), and E is the centre of a circle that shall pass through the angular points A, B, C, D, (f). Wherefore, a circle has been described about a given square; which was to be done. Recite (a) p. 8, 1; (d) ax. 7, 1; (b) p. 9, 1; (c) def. 30, ax. 10, 1; (f) p. 9, 3. 10 P. To describe an isosceles triangle (ALK), having each of the angles (L, K) at the base double of the angle (A) contained by the equal sides (AL, AK). CNB E A F K L Constr. With any radius AB, and centre A, describe the circle BDLK (a); produce BA to meet the circumference in L (6); upon AB describe the square ABCD (c); bisect AB in E (d), and join ED; make EF equal to ED (e); place D AF in the circle from L to K (f); join AK, FK (g); about the triangle AFK describe a circle (h). Argument. The radius AL is so divided in F, that the rectangle ALX LF equals the square of AF (i); but AF is equal to LK (k), which meets the circle AFK in the point K, and AL cuts the same circle in F (1); therefore the square of LK equals the rectangle ALXLF, and LK touches the circle AFK (m). Now ALK is an isosceles triangle, because AL, AK are equal radii (n); therefore the angles ALK, AKL are equal (o). Again, because LK touches the circle AFK, and from the point of contact KF is drawn in the circle, the angle FKL is equal to FAK in the alternate segment (p); but the angle AKL equals FKL and FKA, or FAK and FKA together: and the exterior angle KFL equals the same FAK and FKA (q); therefore KFL equals AKL, or ALK (r), and KFL is an isosceles triangle, in which KL and KF are equal (n): but KL is made equal to AF (b); therefore KF equals AF (r), and AKF is an isosceles triangle (n), having the angles FKA, FAK equal (0); but AKL, or ALK proves equal to both, or double the angle FAK, or LAK. Wherefore, an isosceles triangle has been described, &c. Q. E. F. 11 P. To inscribe an equilateral and equiangular pen tagon in a given circle (ABCDE). Describe an isosceles triangle FGH (a), whose equal sides contain an angle F, half as great as the angle G, or H, at the base. Then in the circle inscribe a triangle ACD, equiangular to FGH; so that the angle C, or D, shall be double the angle A (1). Bisect the angles ACD, C ADC by the chords CE, DB (c); and join AB, BC, CD, DE, EA. ABCDE is the pentagon required. Because each of the angles ACD, ADC is double of the angle CAD; and are bisected by the chords EC, BD, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another: but equal angles stand upon equal arcs (d); therefore the five arcs AB, BC, CD, DE, EA are equal to each other. Again, equal arcs subtend equal chords (e); therefore the five chords AB, BC, CD, DE, EA are equal to each other: wherefore the pentagon is equilateral. It is also equiangular: because the arc AB equals the arc DE; add to each BCD; the sum ABCD is equal to the sum BCDE: and the angle AED stands on the are ABCD; and the angle BAE on the arc BCDE; therefore the angles AED, BAE are equal (f); for the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED; wherefore the pentagon is equiangular. And thus an equilateral and equiangular pentagon has been inscribed in a given circle: which was to be done. Recite (a) p. 10, 4; (b) p. 2, 4; (c) p. 9, 1; (d) p. 26, 3; (e) p. 29, 3; (f) p. 27, 3 12 P. To describe an equilateral and equiangular pentagon about a given circle (ABCDE). Constr. Let A, B, C, D, E, be the angular points of an inscribed pentagon, as in the last (a); so that the arcs AB, BC, CD, DE, EA are H equal: then through the points A, B, C, D, E, draw the tangents GH, HK, KL, LM, MG (6). If these tangents be equal, and their angles B equal, the required pentagon is described about the circle. Find the centre F, and draw the radii FB, FC, FD; join FK, FL. G A E M F D C L K Argument. In the triangles FBK, FCK, the angles at B, C are right angles (c); and the common side FK subtends them: therefore the squares of FB, BK, and also of FC, CK are equal to the square of FK (d), and therefore equal to each other (e). But the squares of the equal radii FB, FC, are equal; therefore the remaining squares of BK, CK are equal; and BK is equal to CK (f). For the same reason CL and DL are equal: therefore the triangles FBK, FCK, FDL, FCL have two sides in each equal to two sides in every other, |