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124. The three tangents will be found to be perpendicular to the sides of the triangle formed by joining the centers of the three circles. 125. With center A and any radius less than the radius of either of the equal circles, describe the third circle intersecting them in C and D. Join BC, CD, and prove BC and CD to be in the same straight line.
126. Let ABC be the triangle required; BC the given base, BD the given difference of the sides, and BAC the given vertical angle. Join CD and draw AM perpendicular to CD. Then MAD is half the vertical angle and AMD a right angle: the angle BDC is therefore given, and hence D is a point in the arc of a given segment on BC. Also since BD is given, the point D is given, and therefore the sides BA, AC are given. Hence the synthesis.
127. Let ABC be the required triangle, AD the line bisecting the vertical angle and dividing the base BC into the segments BD, DC. About the triangle ABC describe a circle and produce AD to meet the circumference in E, then the arcs BE, EC are equal.
128. Analysis. Let ABC be the triangle, and let the circle ABC be described about it: draw AF to bisect the vertical angle BAC and meet the circle in F, make AV equal to AC, and draw CV to meet the circle in T; join TB and TF, cutting AB in D; draw the diameter FS cutting BC in R, DR cutting AF in E; join AS, and draw AK, AH perpendicular to FS and BC. Then shew that AD is half the sum, and DB half the difference of the sides AB, AC. Next, that the point F in which AF meets the circumscribing circle is given, also the point E where DE meets AF is given. The points A, K, R, E are in a circle, Euc. 1. 22. Hence, KF. FRAF.FE, a given rectangle; and the segment KR, which is equal to the perpendicular AH, being given, RF itself is given. Whence the construction.
129. On AB the given base describe a circle such that the segment AEB shall contain an angle equal to the given vertical angle of the triangle. Draw the diameter EMD cutting AB in M at right angles. At D in ED, make the angle EDC equal to half the given difference of the angles at the base, and let DC meet the circumference of the circle in C. Join CA, CB; ABC is the triangle required. For, make CF equal to CB, and join FB cutting CD in G.
130. Let ABC be the triangle, AD the perpendicular on BC. With center A, and AC the less side as radius, describe a circle cutting the base BC in E, and the longer side AB in G, and BA produced in F, and join AE, EG, FC. Then the angle GFC being half the given angle, BAC is given, and the angle BEG equal to GFC is also given. Likewise BE the difference of the segments of the base, and BG the difference of the sides, are given by the problem. Wherefore the triangle BEG is given (with two solutions). Again, the angle EGB being given, the angle AGE, and hence its equal AEG is given; and hence the vertex A is given, and likewise the line AE equal to AC the shortest side is given. Hence the construction.
131. Let ABC be the triangle, D, E the bisections of the sides AC, AB. Join CE, BD intersecting in F. Bisect BD in G and join EG. Then EF, one-third of EC is given, and BG one-half of BD is also given. Now EG is parallel to AC; and the angle BAC being given, its equal opposite angle BEG is also given. Whence the segment of the circle containing the angle BEG is also given. Hence F is a given point, and FE a given line, whence E is in the circumference of the given circle about whose radius is FE. Wherefore E being in two given circles, it is itself their given intersection.
132. Of all triangles on the same base and having equal vertical angles, that triangle will be the greatest whose perpendicular from the vertex on the base is a maximum, and the greatest perpendicular is that which bisects the base. Whence the triangle is isosceles.
133. Let AB be the given base and ABC the sum of the other two sides; at B draw BD at right angles to AB and equal to the given altitude, produce BD to E making DE equal to BD. With center A and with radius AC describe the circle CFG, draw FO at right angles to BE and find in it the center O of the circle which passes through B and E and touches the former circle in the point F. The centers A, O being joined and the line produced, will pass through F. Join OB. Then AOB is the triangle required.
Since the area and bases of the triangle are given, the altitude is given. Hence the problem is-given the base, the vertical angle and the altitude, describe the triangle.
135. Apply Euc. III. 27.
136. The fixed point may be proved to be the center of the circle. 137. Let the line which bisects any angle BAD of the quadrilateral, meet the circumference in E, join EC, and prove that the angle made by producing DC is bisected by EC.
138. Draw the diagonals of the quadrilateral, and by Euc. III. 21, I. 29. 139. From the center draw lines to the angles: then Euc. III. 27.
140. The centers of the four circles are determined by the intersection of the lines which bisect the four angles of the given quadrilateral. Join these four points, and the opposite angles of the quadrilateral so formed are respectively equal to two right angles.
141. Let ABCD be the required trapezium inscribed in the given circle (fig. Euc. III. 22.) of which AB is given, also the sum of the remaining three sides and the angle ADC. Since the angle ADC is given, the opposite angle ABC is known, and therefore the point C and the side BC. Produce AD and make DE equal to DC and join EC. Since the sum of AD, DC, CB is given, and DC is known, therefore the sum of AD, DC is given, and likewise AC, and the angle ADC. Also the angle DEC being half of the angle ADC is given. Whence the segment of the circle which contains AEC is given, also AE is given, and hence the point E, and consequently the point D. Whence the construction.
142. Let ADBC be the inscribed quadrilateral; let AC, BD produced meet in O, and AB, CD produced meet in P, also let the tangents from O, P meet the circles in K, H respectively. Join OP, and about the triangle PAC describe a circle cutting PO in G and join AG. Then A, B, G, O may be shewn to be points in the circumference of a circle. Whence the sum of the squares on OH and PK may be found by Euc. III. 36, and shewn to be equal to the square on OP.
143. This will be manifest from the equality of the two tangents drawn to a circle from the same point.
144. Apply Euc. III. 22.
145. A circle can be described about the figure AECBF.
146. Apply Euc. III. 22, 32.
147. Apply Euc. III. 21, 22, 32.
148. Apply Euc. 11. 20, and the angle BAD will be found to be double of the angles CBD and CDB together.
149. Let ABCD be the given quadrilateral figure, and let the angles at A, B, C, D be bisected by four lines, so that the lines which bisect the angles A and B, B and C, C and D, D and A, meet in the points a, b, c, d, respectively. Prove that the angles at a and c, or at b and d, are to
her equal to two right angles.
150. Apply Euc. 11. 22.
151. Join the center of the circle with the other extremity of the line perpendicular to the diameter.
152. Let AB be a chord parallel to the diameter FG of the circle, fig. Theo. 1, p. 160, and H any point in the diameter. Let HA and HB be joined. Bisect FG in O, draw OL perpendicular to FG cutting AB in K, and join HK, HL, OA. Then the square on HA and HF may be proved equal to the squares on FH, HG by Theo. 3, p. 114; Euc. 1. 47; Euc. 11. 9.
153. Let A be the given point (fig. Euc. 111. 36, Cor.) and suppose AFC meeting the circle in F, C, to be bisected in F, and let AD be a tangent drawn from A. Then 2. AF2 AF. AC AD2, but AD is given, hence also AF is given. To construct. Draw the tangent AD. On AD describe a semicircle AGD, bisect it in G; with center A and radius AG, describe a circle cutting the given circle in F. Join AF and produce it to meet the circumference again in C.
154. Let the chords AB, CD intersect each other in E at right angles. Find F the center, and draw the diameters HEFG, AFK and join AC, CK, BD. Then by Euc. 11. 4. 5; 111. 35.
155. Let E, F be the points in the diameter AB equidistant from the center O; CED any chord; draw OG perpendicular to CED, and join FG, OC. The sum of the squares on DF and FC may be shewn to be equal to twice the square on FE and the rectangle contained by AE, EB by Euc. 1. 47; 11. 5; 111. 35.
156. Let the chords AB, AC be drawn from the point A, and let a chord FG parallel to the tangent at A be drawn intersecting the chords AB, AC in D and E, and join BC. Then the opposite angles of the quadrilateral BDEC are equal to two right angles, and a circle would circumscribe the figure. Hence by Euc. 1. 36.
157. Let the lines be drawn as directed in the enunciation. Draw the diameter AE and join CE, DE, BE; then AC2+AD' and 2. AB* may be each shewn to be equal to the square on the diameter. 158. Let QOP cut the diameter AB in O. From C the center draw CH perpendicular to QP. Then CH is equal to OH, and by Euc. 11. 9, the squares on PO, OQ are readily shewn to be equal to twice the square on CP.
159. From P draw PQ perpendicular on AB meeting it in Q. Join AC, CD, DB. Then circles would circumscribe the quadrilaterals ACPQ and BDPQ, and then by Euc. III. 36.
160. Describe the figure according to the enunciation; draw AE the diameter of the circle, and let P be the intersection of the diagonals of the parallelogram. Draw EB, EP, EC, EF, EG, EH. Since AE is a diameter of the circle, the angles at F, G, H, are right angles, and EF, EG, EH are perpendiculars from the vertex upon the bases of the triangles EAB, EAC, EAP. Whence by Euc. 11. 13, and theorem 3, page 114, the truth of the property may be shewn.
161. If FA be the given line (fig. Euc. 11. 11), and if FA be produced to C; AC is the part produced which satisfies the required conditions.
162. Let AD meet the circle in G, H, and join BG, GC. Then BGC is a right-angled triangle and GD is perpendicular to the hypotenuse, and the rectangles may be each shewn to be equal to the square on BG. Euc. III. 35; II. 5; 1. 47. Or, if EC be joined, the quadrilateral figure ADCE may be circumscribed by a circle. Euc. III. 31, 22, 36, Cor.
163. On PC describe a semicircle cutting the given one in E, and draw EF perpendicular to AD; then F is the point required.
164. Let AB be the given straight line. Bisect AB in C and on AB as a diameter describe a circle; and at any point D in the circumference, draw a tangent DE equal to a side of the given square; join DC, EC, and with center C and radius CE describe a circle cutting AB produced in F. From F draw FG to touch the circle whose center is C in the point G.
165. Let AD, DF be two lines at right angles to each other, O the centre of the circle BFQ; A any point in AD from which tangents AB, AC are drawn; then the chord BC shall always cut FD in the same point P, wherever the point A is taken in AD. Join AP; then BAC is an isosceles triangle,
and FD. DE + AD2 = AB2 = BP. PC + AP2 = BP. PC + AD2 + DP2, wherefore BP.PC FD. DE - DP. The point P, therefore, is independent of the position of the point A ; and is consequently the same for all positions of A in the line AD.
166. The point E will be found to be that point in BC, from which two tangents to the circles described on AB and CD as diameters, are equal, Euc. III. 36.
167. If AQ, A'P' be produced to meet, these lines with AA' form a right-angled triangle, then Euc. I. 47.
GEOMETRICAL EXERCISES ON BOOK IV.
1. LET AB be the given line. Draw through C the center of the given circle the diameter DCE. Bisect AB in F and join FC. Through A, B draw AG, BH parallel to FC and meeting the diameter in G, H: at G, H draw GK, HL perpendicular to DE and meeting the circumference in the points K, L ; join KL; then KL is equal and parallel to AB.
2. Trisect the circumference and join the center with the points of trisection.
3. See Euc. IV. 4, 5.
4. Let a line be drawn from the third angle to the point of intersection of the two lines; and the three distances of this point from the angles may be shewn to be equal.
5. Let the line AD drawn from the vertex A of the equilateral triangle, cut the base BC, and meet the circumference of the circle in D. Let DB, DC be joined: AD is equal to DB and DC. If on DA, DE be taken equal to DB, and BE be joined; BDE may be proved to be an equilateral triangle, also the triangle ABE may be proved equal to the triangle CBD.
The other case is when the line does not cut the base.
6. Let a circle be described upon the base of the equilateral triangle, and let an equilateral triangle be inscribed in the circle. Draw a diameter from one of the vertices of the inscribed triangle, and join the other extremity of the diameter with one of the other extremities of the sides of the inscribed triangle. The side of the inscribed triangle may then be proved to be equal to the perpendicular in the other triangle.
7. The line joining the points of bisection, is parallel to the base of the triangle and therefore cuts off an equilateral triangle from the given triangle. By Euc. III. 21; 1. 6, the truth of the theorem may be shewn.
8. Let a diameter be drawn from any angle of an equilateral tri
angle inscribed in a circle to meet the circumference. It may be proved that the radius is bisected by the opposite side of the triangle.
9. Let ABC be an equilateral triangle inscribed in a circle, and let AB'C' be an isosceles triangle inscribed in the same circle, having the same vertex A. Draw the diameter AD intersecting BC in E, and B'C' in E', and let B'C' fall below BC. Then AB, BE, and AB', B'E', are respectively the semi-perimeters of the triangles. Draw B'F perpendicular to BC, and cut off AH equal to AB, and join BH. If BF can be proved to be greater than B'H, the perimeter of ABC is greater than the perimeter of AB'C'. Next let B'C' fall above BC.
10. The angles contained in the two segments of the circle, may be shewn to be equal, then by joining the extremities of the arcs, the two remaining sides may be shewn to be parallel.
11. It may be shewn that four equal and equilateral triangles will form an equilateral triangle of the same perimeter as the hexagon, which is formed by six equal and equilateral triangles.
12. Let the figure be constructed. By drawing the diagonals of the hexagon, the proof is obvious.
13. By Euc. 1. 47, the perpendicular distance from the center of the circle upon the side of the inscribed hexagon may be found.
14. The alternate sides of the hexagon will fall upon the sides of the triangle, and each side will be found to be equal to one-third of the side of the equilateral triangle.
15. A regular duodecagon may be inscribed in a circle by means of the equilateral triangle and square, or by means of the hexagon. The area of the duodecagon is three times the square on the radius of the circle, which is the square on the side of an equilateral triangle inscribed in the same circle. Theorem 1, p. 196.
16. In general, three straight lines when produced will meet and form a triangle, except when all three are parallel or two parallel are intersected by the third. This Problem includes Euc. Iv. 5, and all the cases which arise from producing the sides of the triangle. The circles described touching a side of a triangle and the other two sides produced, are called the escribed circles.
17. This is manifest from Euc. II. 21.
18. The point required is the center of the circle which circumscribes the triangle. See the notes on Euc. 11. 20, p. 155.
19. If the perpendiculars meet the three sides of the triangle, the point is within the triangle, Euc. IV. 4. If the perpendiculars meet the base and the two sides produced, the point is the center of the escribed circle.
20. This is manifest from Euc. III. 11, 18.
21. The base BC is intersected by the perpendicular AD, and the side AC is intersected by the perpendicular BE. From Theorem 1. p. 160; the arc AF is proved equal to AE, or the arc FE is bisected in A. In the same manner the arcs FD, DE, may be shewn to be bisected in BC.
22. Let ABC be a triangle, and let D, E be the points where the inscribed circle touches the sides AB, AC. Draw BE, CD intersecting each other in O. Join AO, and produce it to meet BC in F. Then Fis the point where the inscribed circle touches the third side BC. If F be not the point of contact, let some other point G be the point of contact. Through D draw DH parallel to AC, and DK parallel to BC. By the similar triangles, CG may be proved equal to CF, or G the point of contact coincides with F, the point where the line drawn from A through O meets BC.