AM, AN, and on AP or AP produced, take AD equal to the given perpendicular, and through D draw BC parallel to MN meeting AM, AN, or these lines produced. Then ABC shall be the triangle required.

85. Let PAQ be the given angle, bisect the angle A by AB, in AB find D the center of the inscribed circle, and draw DC perpendicular to AP. In DB take DE such that the rectangle DE, DC is equal to the given rectangle. Describe a circle on DE as diameter meeting AP in F, G; and AQ in F, G'. Join FG', and AFG will be the triangle. Draw DH perpendicular to FG and join GD. By Euc. vI. C, the rectangle FD, DG is equal to the rectangle ED, DK or CD, DE.

86. On any base BC describe a segment of a circle BAC containing an angle equal to the given angle. From D the middle point of BC draw DA to make the given angle ADC with the base. Produce AD to E so that AE is equal to the given bisecting line, and through E draw FG parallel to BC. Join AB, AC and produce them to meet FG in F and G. 87. Employ Theorem 70, p. 310, and the construction becomes obvious.

88. Let AB be the given base, ACB the segment containing the vertical angle; draw the diameter AB of the circle, and divide it in E, in the given ratio; on AE as a diameter, describe a circle AFE; and with center B and a radius equal to the given line, describe a circle cutting AFE in F. Then AF being drawn and produced to meet the circumscribing circle in C, and CB being joined, ABC is the triangle required. For AF is to FC in the given ratio.

89. The line CD is not necessarily parallel to AB. Divide the base AB in C, so that AC is to CB in the ratio of the sides of the triangle. Then if a point E in CD can be determined such that when AE, CE, EB, are joined, the angle AEB is bisected by CE, the problem is solved. 90. Let ABC be any triangle having the base BC. On the same base describe an isosceles triangle DBC equal to the given triangle. Bisect BC in E, and join DE, also upon BC describe an equilateral triangle. On FD, FB, take EG to EH as EF to FB: also take EK equal to EH and join GH, GK; then GHK is an equilateral triangle equal to the triangle ABC.

91. Let ABC be the required triangle, BC the hypotenuse, and FHKG the inscribed square: the side HK being on BC. Then BC may be proved to be divided in H and K, so that HK is a mean proportional between BH and KC.

92. Let ABC be the given triangle. On BC take BD equal to one of the given lines, through A draw AE parallel to BC. From B draw BE to meet AE in E, and such that BE is a fourth proportional to BC, BD, and the other given line. Join EC, produce BE to F, making BF equal to the other given line, and join FD: then FBD is the triangle required.

93. By means of Euc. vI. C, the ratio of the diagonals AC to BD may be found to be as AB. AD + BC. CD to AB. BE+ AD.DC, figure, Euc. vi. D.

94. This property follows directly from Euc. vi. C.

95. Let ABC be any triangle, and DEF the given triangle to which the inscribed triangle is required to be similar. Draw any line de terminated by AB, AC, and on de towards AC describe the triangle def similar to DEF, join Bf, and produce it to meet AC in F. Through F draw F'D' parallel to fd, F'E' parallel to fe, and join D'E', then the triangle D'E'F' is similar to DEF.

96. The square inscribed in a right-angled triangle which has one of its sides coinciding with the hypotenuse, may be shewn to be less than that which has two of its sides coinciding with the base and perpendicular. 97. Let BCDE be the square on the side BC of the isosceles triangle ABC. Then by Euc. vi. 2, FG is proved parallel to ED or BC.

98. Let AB be the base of the segment ABD, fig. Euc. 111. 30. Bisect AB in C, take any point E in AC and make CF equal to CE: upon EF describe a square EFGH: from C draw CG and produce it to meet the arc of the segment in K.

99. Take two points on the radii equidistant from the center, and on the line joining these points, describe a square; the lines drawn from the center through the opposite angles of the square to meet the circular arc, will determine two points of the square inscribed in the sector.

100. Let ABCDE be the given pentagon. On AB, AE take equal distances AF, AG, join FG, and on FG describe a square FGKH. Join AH and produce it to meet a side of the pentagon in L. Draw LM parallel to FH meeting AE in M. Then LM is a side of the inscribed square.

101. Let ABC be the given triangle. base BC an angle equal to one of the given Draw AE parallel to BC and take AD to sides. Join BE cutting AC in F.

Draw AD making with the angles of the parallelogram. AE in the given ratio of the

102. The locus of the intersections of the diagonals of all the rectangles inscribed in a scalene triangle, is a straight line drawn from the bisection of the base to the bisection of the shorter side of the triangle. 103. This parallelogram is one half of the square in the circle.

104. Analysis. Let ABCD be the given rectangle, and EFGH that to be constructed. Then the diagonals of EFGH are equal and bisect each other in P the center of the given rectangle. About EPF describe a circle meeting BD in K, and join KE, KF. Then since the rectangle EFGH is given in species, the angle EPF formed by its diagonals is given; and hence also the opposite angle EKF of the inscribed quadrilateral PEKF is given. Also since KP bisects that angle, the angle PKE is given, and its supplement BKE is given. And in the same way, KF is paralled to another given line; and hence EF is parallel to a third given line. Again, the angle EPF of the isosceles triangle EPF is given; and hence the quadrilateral EPFK is given in species.

105. In the figure Euc. III. 30; from C draw CE, CF making with CD, the angles DCE, DCF each equal to the angle CDA or CDB, and meeting the arc ADB in E and F. Join EF, the segment of the circle described upon EF and which passes through C, will be similar to ADB.

106. The square inscribed in the circle may be shewn to be equal to twice the square on the radius; and five times the square inscribed in the semicircle to four times the square on the radius.

107. The three triangles formed by three sides of the square with segments of the sides of the given triangle, may be proved to be similar. Whence by Euc. vi. 4, the truth of the property.

108. By constructing the figure, it may be shewn that twice the square inscribed in the quadrant is equal to the square on the radius, and that five times the square inscribed in the semicircle is equal to four times the square on the radius. Whence it follows that, &c.

109. By Euc. 1. 47, and Euc. vI. 4, it may be shewn, that four times the square on the radius is equal to fifteen times the square on one of the equal sides of the triangle.

110. Constructing the figure, the right-angled triangles SCT, ACB

may be proved to have a certain ratio, and the triangles ACB, CPM in the same way, may be proved to have the same ratio.

111. Let BA, AC be the bounding radii, and D a point in the arc of a quadrant. Bisect BAC by AE, and draw through D, the line HDGP perpendicular to AE at G, and meeting AB, AC, produced in H, P. From H draw HM to touch the circle of which BC is a quadrantal arc; produce AH, making HL equal to HM, also on HA, take HK equal to HM. Then K, L, are the points of contact of two circles through D which touch the bounding radii, AB, AC.

Join DA. Then, since BAC is a right angle, AK is equal to the radius of the circle which touches BA, BC in K, K'; and similarly, AL is the radius of the circle which touches them in L, L'. Also, HAP being an isosceles triangle, and AD drawn to the base, AD' is shewn to be equal to AK. KL. Euc. III. 36; 11. 5, Cor.

112. Let E, F, G be the centers of the circles inscribed in the triangles ABC, ADB, ACD. Draw EH, FK, GL perpendiculars on BC, BA, AC respectively, and join CE, EB; BF, FA; CG, GA. Then the relation between R, r, r', or EH, FK, GL may be found from the similar triangles, and the property of right-angled triangles.

113. The two hexagons consist each of six equilateral triangles, and the ratio of the hexagons is the same as the ratio of their equilateral triangles.

114. The area of the inscribed equilateral triangle may be proved to be equal to half of the inscribed hexagon, and the circumscribed triangle equal to four times the inscribed triangle.

115. The pentagons are similar figures, and can be divided into the same number of similar triangles. Euc. vi. 19.

116. Let the sides AB, BC, CA of the equilateral triangle ABC touch the circle in the points D, E, F, respectively. Draw AE cutting the circumference in G; and take O the center of the circle and draw OD: draw also HGK touching the circle in G. The property may then be shewn by the similar triangles AHG, AOD.

GEOMETRICAL EXERCISES ON BOOK XI.
HINTS, &c.

3. Let AD, BE be two parallel straight lines, and let two planes ADFC, BEFC pass through AD, BE, and let CF be their common intersection, fig. Euc. x1, 10. Then CF may be proved parallel to BE and AD.

4. This theorem is analogous to Euc. XI. 8. Let two parallel lines AC, BD meet a plane in the points A, B. Take AC equal to BD and draw CE, DF, perpendiculars on the plane, and join AE, BF. Then the angles CAE, DBF, are the inclinations of AC, BD to the plane, Euc. XI. def. 5, and these angles may be proved to be equal.

5. Let AB, CD be parallel straight lines, and let perpendiculars be drawn from the extremities of AB, CD on any plane, and meet it in the points A', B', C', D'. Draw A'B', C'D'; these are the projections of AB, CD on the plane, and may be proved to be parallel.

6. Draw the figure, the proof offers no difficulty.

Let AB, AČ drawn from the point A, and A'B', A'C' drawn from

the point A', in two parallel planes, make equal angles with a plane EF passing through AA', and perpendicular to the planes BAC, B'A'C'. Let AB in the plane ABC be parallel to A'B' in the plane A'B'C': then AC may be proved to be parallel to A'C'.

8. The plane must be drawn through the given line so that the plane and the other given line may be equally inclined to a third plane. 9. The required plane must be drawn through the given point so 23 to have the same inclination to a third plane, as the plane which passes through the two given lines.

10. From the point A let AB be drawn perpendicular to a plane, and AC perpendicular to a given line CD in a plane: join BC, then BC is at right angles to CD. For AB, BC, CD may be considered as three consecutive edges of a rectangular parallelopiped, and AC the diagonal of one face.

11. In the triangle BCD in which BE is drawn from the vertex to a point E in the base CD; it may be proved that the difference of the squares on the sides BC, BD is equal to the difference of the squares on the segments CE, ED of the base. By the converse of Theo. 149, p. 83.

12. Let BC be the common intersection of the two planes ABCD, EFGH which are inclined to each other at any angle. From K at any point in the plane ABCD, let KL be drawn perpendicular to the plane EFGHI, and KM perpendicular to BC, the line of intersection of the two planes. Join LM, and prove that the plane which passes through KL, KM is perpendicular to the line BC.

13. About the given line let a plane be made to revolve, till it passes through the given point. The perpendicular drawn in this plane from the given point upon the given line is the distance required.

14. Through any point in the first line draw a line parallel to the second; the plane through these is parallel to the second line. Through the second line draw a plane perpendicular to the fore-named plane cutting the first line in a point. Through this point draw a perpendicular in the second plane to the first, and it will be perpendicular to both lines.

15. Through any point draw perpendiculars to both planes; the plane passing through these two lines will fulfil the conditions required.

16. From the points where the lines meet the planes, draw two lines perpendicular to the intersection of the planes.

17. Let AB, AC in one of the planes make equal angles with DE the line of the intersection of the planes. Let AB be equal to AC. Draw BF, CG perpendiculars on the other plane, and draw FA, GA in that plane, and prove the angle BAF equal to the angle CAG.

18. If the intersecting plane be perpendicular to the three straight lines; by joining the points of their intersection with the plane, the figure formed will be an equilateral triangle. If the plane be not perpendicular, the triangle will be isosceles.

19. Let the straight lines intersect in A, and let a plane be drawn cutting the three given lines in the points B, C, D, and the fourth in E. 20. This will appear from Euc. 1. 19.

21: Let S be the proposed solid angle, in which the three plane angles ASB, ASC, BSC are known, it is required to find the angle contained by two of these planes, such as ASB, ASC. On a plane make the angles B'SA, ASC, B"SC equal to the angles BSA, ASC, BSC in the solid figure; take B'S and B'S each equal to BS in the solid figure; from the points B', and B" at right angles to SA and SC draw B'A and B"C, which will intersect each other at the point 0. From () as a center, with radius

AB' describe the semicircle B'bE; at the point O, erect Ob perpendicular to B'E and meeting the circumference in b; join Ab: the angle EAb will be the required inclination of the two planes ASC, ASB in the solid angle. (Legendre's Geometry, translated by Sir David Brewster, pp. 125, &c.)

22. Let ASC, ASB (same figure as in 21) be the two given plane angles; and suppose for a moment that CSB" is the third angle required; then employing the same construction as in the foregoing problem, the angle included between the planes of the two first, the inclination of these planes would be EAb. Now as EAb can be determined by means of CSB", the other two being given, so likewise may CSB" be determined by means of EAb, which is just what the problem requires.

Having taken SB' at pleasure, upon SA let fall the indefinite perpendicular B'E; make the angle EAb equal to the inclination of the two given planes; from the point b, where the side Ab meets the circle described from the center A with the radius AB', draw 60 perpendicular to AE; from the point O, at right angles to SC draw the indefinite line OCB"; make SB" equal to SB'; the angle CSB" will be the third plane angle required. (Legendre's Geometry, translated by Sir David Brewster, pp. 127, &c.)

23. Let the three lines meet in the point A, and let a plane intersect them in the points B, C, D, so that AB, AC, AD are equal to one another. Describe a circle about the triangle BCD, and let O be the center; the line AO is perpendicular to the plane BCD.

24. This may be readily proved by Euc. XI. 17.

25. Construct the figure, and it will be found that the angle between the diagonal and one side of the cube measures the inclination of the two planes.

26. The diagonal plane of a cube is at right angles to two of the faces of the cube, and makes angles, each equal to half a right angle with the other four faces.

27. Let a rectangular parallelogram ABCD, be formed by four squares, each equal to a face of the given cube, and let EF, GH, KL, be the lines of division of the four squares. Let BD the diagonal of ABCD, cut EF in M: the square on BM to the square on AB is as 17 to 16. Let BG the diagonal of ABHG cut EF in N; the square on BN is to the square on AB, as 20 is to 16; hence there is some square between that on BM and BN which bears to the square on AB, the ratio of 18 to 16, or of 9 to 8.

The following addition may be easily proved. If six edges of a cube taken in order round the figure, be bisected, and the points of bisection be joined in succession, these six lines will form a regular hexagon.

28. From the six points out of the perpendicular, draw perpendiculars to the plane, and join the points where the perpendiculars meet the plane. 29. This is to shew that the square on the diagonal of a rectangular parallelopiped is equal to the sum of the squares on its three edges. 30. This theorem is analogous to the corresponding theorem respecting a rectangular parallelogram.

The axis of a parallelopiped must not be confounded with its diagonal. 31. Let the figure be described in a similar manner to that of Theorem 2, page 337: by employing Euc. 11. 12, 13, instead of Euc. 1. 47, the truth of the theorem may be proved.

32. Describe a circle passing through the three given points, and from the center draw a line perpendicular to its plane. Then every point in this perpendicular fulfils the conditions required.