perpendicular to any side BC (d), and from Das a center, and with the distance DF describe a circle EFG which shall be inscribed in the given triangle. E B (d) I. 12. DEMONSTRATION. From D draw DE and DG perpendicular to AB and AC. Then the angle ABC being bisected by DB (e), the angles EBD and FBD are equal, and the angles DEB and DFB being both right angles (f) are also equal, therefore the triangles EBD and FBD have two angles of the one respectively, equal to two angles of the other, and the side BD common to both, and therefore their other sides ED and FD are equal (g). In the same manner it may be shown that GD is equal to FD; therefore the three lines ED, FD, and GD are equal (h), and therefore the circle described from the center D, with the radius DF, passes through the points E and G, and because the angles at F, É, and G are right angles, the lines BC, AB, and AC are tangents to the circle (i); therefore the circle FEG is inscribed in the given triangle (k). SCHOLIUM. The above proposition is only a particular case of the more general problem," To describe a circle touching three given straight lines." 1o. If the three given lines are parallel to each COROLLARY 1. The straight lines bisecting the three angles of a triangle meet in the center of the inscribed circle. COROLLARY 2. A triangle is equal in area to the rectangle under the radius of the inscribed circle, and half the sum of the three sides or perimeter of the triangle. A C /F. D B For the area of the whole triangle ABC is equal to the areas of the three triangles AEB, BEC, and AEC, and the area of each of these triangles is respectively equal to that of the rectangle, under the radius and half the sides AB, BC, and AC. Ꭺ. E B PROPOSITION V. PROBLEM. To circumscribe a circle about a given triangle (ABC). SOLUTION. The three angular points, A, B, and C, of the triangle, are not in the same straight line, therefore a circle may be described passing through them in the manner demonstrated in the theorem attached to III. 1. 2. If the center F fall within the triangle all its angles are acute, for each of them is in a segment greater than a semicircle. If the center be in any side of the triangle, the angle opposite that side is a right angle, because it is in a semicircle. And if the center fall without the triangle, the angle opposite to the side which is nearest to the center is an obtuse angle, because it is in a segment greater than a semicircle. 3. The two following propositions are here introduced, in order to simplify the demonstration of several of the subsequent problems. PROPOSITION V. A. THEOREM.-If a rectilineal figure (ABCDE) be equilateral and equiangular, [1] it may have one circle cir- B cumscribed about it, [2] and another inscribed in it; [3] and the same point is the center of both circles. CONSTRUCTION. Bisect the angles BCD and CDE (a), by the straight lines CF and DF, then because the angles FCD and FDC are together less than two right angles, therefore CF and DF will meet, if produced far enough (b), let them meet in F. Join BF, and from F draw GF and HF respectively B perpendicular to BC and CD (c). H D E (b) Theor. attached to I. 29. DEMONSTRATION. [1.] In the triangle FCD, the angles FCD and FDC are equal, being the halves of equal angles, therefore the opposite sides CF and DF are equal (d). Also in the triangles FBC and FDC, the side BC is equal to CD (e), the side CF common to both, and the angle FCB equal to FCD (f), therefore the side BF is equal to DF (g). In the same manner it may be shown that the straight lines from F to the other angles A and E are equal to DF, and therefore a circle described from F as a center, with the radius DF, will pass through all the angular points, and circumscribe the rectilineal figure ABCDE. (c) I. 12. [2.] Because BC and CD are equal chords of the circumscribed circle, they are equally distant from its center (h), that is, GF is equal to HF; in the same manner it may be shown that the perpendiculars drawn from F to the other sides AB, AE, and DE are all equal to HF, and therefore that a circle described from F as a center, with the radius HF, will touch all the sides of the rectilineal figure ABCDE, and be inscribed in it. [3.] It is evident that the same point F is the center of both the circumscribed and inscribed circles. PROPOSITION V. B. THEOREM.-If any equilateral and equiangular rectilineal figure (ABCDE) be inscribed in a circle, tangents to the circle, drawn through the angular points, will form an equilateral and equiangular figure of the same number of sides, circumscribed about the circle. DEMONSTRATION. Because the chords AB and BC, &c., are equal, their arcs are also equal (a), and the angles FAB, FBA, GBC, GCB, &c., at the circumference standing on these arcs are also equal (b). Therefore in the triangles ABF, G BCG, GHD, &c., the sides AF and BF, BG, GC, CH, &c., are all equal, and the angles F, G, H, &c., are also all equal (c), therefore the rectilineal figure, FGHIK, circumscribed about the circle, is equilateral and equiangular. B F A II K (a) III. 28. E PROPOSITION VI. PROBLEM. To inscribe a square in a given circle (ABCD). SOLUTION. Draw the diameters AC, BD, at right angles to each other; and join AB, BC, CD, and DA, then ABCD is the square required. B E D (a) I. 4. DEMONSTRATION. Because in the triangles BEA and AED, BE and ED are equal, AE common to both, and at right angles to BD, the base AB is equal to AD (a); and in the same manner it may be shown that each of the other sides, DC and BC are equal to AB, and therefore that the quadrilateral figure ABCD is equilateral. But the straight line BD being a diameter, ABD is a semicircle, and therefore the angle BAD is a right angle (b), and the quadrilateral figure ABCD is a square. PROPOSITION VII. PROBLEM. To circumscribe a square about a given circle (ABCD). SOLUTION. Inscribe a square in the circle ABCD E A B DEMONSTRATION. Because the tangents, EH, EF, FG, and GH are drawn through the angular r points of a square inscribed in a circle, therefore they form a square EFGH, circumscribed about the same circle (c). C (a) IV. 6. (b) III. 17. (c) IV. 5 B. G D COROLLARY. If a square is circumscribed about a circle, it is evidently equal in area to twice the square inscribed in the circle. SCHOLIUM. It is evident that a square is the only right-angled parallelogram which can be circumscribed about a circle, but that either a square or rectangle may be inscribed in it. PROPOSITION VIII. PROBLEM.-To inscribe a circle in a given square (ABCD). pa SOLUTION. Bisect each of the sides AB, AD, in A the points F and E (a); through F draw EK rallel to AD, and through E draw EH parallel to AB (b); the circle EFHK, described from the center G, with the radius EG, is inscribed in the given square. F B (a) I. 10. (b) I. 31. (c) Ax. 7. (d) Solution. (e) I. 34. (ƒ) III. 16. K DEMONSTRATION. Because AE, ED, AF, and FB are halves of equals, they are all equal to each other (c); and because AG, EK, FH, and GC are parallelograms (d), their opposite sides are equal (e); therefore, EG, FG, HG, and GK are all equal, and the circle described from the center G with the radius EG will pass through the points F, H, and K, and because the angles at E, F, H, and K, are right angles (d), the sides of the square are tangents to the circle EFHK (ƒ); which is therefore inscribed in the given square. PROPOSITION IX. PROBLEM.-To circumscribe a circle about a given square (ABCD). SOLUTION. Join AC and BD, cutting each other in E; the circle described from E as a center with the radius AE will circumscribe the given square. A B E (a) I. 32 B, cor. 2. (6) I. 6. D C DEMONSTRATION. Because the triangle ABD is isosceles, and the angle A a right angle, therefore each of the angles ADB and ABD is half a right angle (a), and in the same manner it may be shown that each of the angles into which the angles of the square are divided by the diagonals is half a right angle; and, therefore, that they are all equal. Then in the triangle, AEB, as the angles A and B are equal, the opposite sides, BE and AE, are equal (b); and in the same manner it may be shown that CE and DE are equal to BE and AE, therefore the four lines, AE, BE, CE, and DE, are equal, and therefore the circle described from the center E with the radius AE passes through the angular points, A, B, C, and D, and is circumscribed about the given square. |