.. DH equal to DF, and GH equal to GF; DEH is the triangle required. For DE: DA :: DF: DB :: DE + DH : DA [+DB, •. DE+DH is equal to the given sum. Also the angle EHD is equal to the two HDF, DFH, i.e. to HDF, DHF, and DEH is equal to the difference between DHF and EDH; :. the difference between EHD and DEH is equal to the sum of HDF and EDH, i.e. to EDF, or the given difference. Also EG : GH :: EG : GF :: AC : CB i. e. in the given ratio. A с G (23.) Given the difference of the angles at the base, the ratio of the sides, and the length of a third proportional to the difference of the segments of the base made by a perpendicular from the vertex and the shorter side ; to construct the triangle. Let ABC be equal to the given difference of the angles at the base ; and take AB : BC in the given ratio of the sides. Join AC; and produce BC to D, so that BD may be to the given third proportional in the ratio CA : CB. Through D draw EDF parallel to AC; . and from B draw BG perpendicular to it; make GF= GD; join BF; EBF is the triangle required. Since DG=GF, and the angles at G are right angles; •. BF = BD, and the angle BFD = BDF. Hence the difference between BFE and BEF is equal to the difference between BDF and BEF, i. e, to EBD or the given difference. аа Also EB : BF :: EB : BD :: AB : BC, i. e, in the given ratio. And DE : BF :: DE : BD :: CA : CB :: (BD=) BF : the given third proportional, whence DE is equal to the given difference of the segments of the base. B G G (24.) Given the base of a right-angled triangle ; to construct it, when parts, equal to given lines, being cut off from the hypothenuse and perpendicular, the remainders have a given ratio. Let AB be equal to the given base. From B draw BC at right angles to it, and equal to the part to be cut off from the perpendicular. Take CD to the given part to be cut off from the hypothenuse, in the given ratio of the remainders ; and from C to DA, draw CE equal to that given part, and produce it. From A draw AF perpendicular to AB, meeting CE produced in F. Draw FG parallel to AB; CFG is the triangle required. Since AF is parallel to CD, the angles AFE, ECD are equal, and the angles at E are equal, .-. the triangles AEF, CED are equiangular, and .:. FE : AF :: CE : CD. And since AG is a parallelogram, AF=BG, .:. FE : BG :: CE : CD, i.e. in the given ratio of the remainders; and FG is equal to the given base; and CB, CE are equal to the given parts to be cut off. sums. H G E (25.) Given one angle of a triangle, and the sums of each of the sides containing it and the third side; to construct the triangle. Let BAC be equal to the given angle; and AB, AC equal to the given Join BC; and draw any line DE parallel to it. Make CF, FG, each equal to BD; join CG, and produce it to H; and draw HI parallel to GF. AHI is the triangle required. Since DG is parallel to BC, and GF to HI, BD : BH (:: CG : CH) :: GF: HI :: CF : and since BD, GF and FC are equal, BH, HI and IC are also equal; whence AH and HI together are equal to AB; and AI, IH are together equal to AC; and HAI is equal to the given angle. B (26.) Given the vertical angle, and the ratio of the sides containing it, as also the diameter of the circumscribing circle ; to construct the triangle. On the given diameter describe a circle ; and from it cut off a segment ACB containing an angle equal to the given vertical angle, Divide the base AB in D, in the given ratio of the sides ; and draw the diameter EF at right angles to AB. Join ED, and let it meet the circumference in C; join AC, CB; ACB is the triangle required. Since AE=EB, the angle ACB is bisected by CE, .: AC : CB :: AD : DB, i.e. in the given ratio. GI/D B (27.) Given the vertical angle, and the radii of the inscribed and circumscribing circles; to construct the triangle. With the given radius describe the circumscribing circle; and from any point A in it take AB, AC each equal to the arc subtending at the centre an angle equal to the given angle. Join BC; and parallel to it, at a distance equal to the radius of the inscribed circle draw EF. Join AB; and from A to EF draw AO=AB, and produce 40 to G. Join CG, BG; GCB is the triangle required. E A For the angle CGB (Eucl. iii. 20.) is equal to the angle at the centre, which stands on AC, and .. is equal to the given angle. Also the angle CGA=AGB. Join BO; and since AB=AO, the angle AOB = ABO; but AOB is equal to the two OBG, OGB; and AGB= ABC, CBO= OBG, and BQ bisects the angle CBG; whence O, the point of intersection of the bisecting lines GA and BO, is the centre of the inscribed circle; and its distance from BC was made equal the given radius. (28.) Given the vertical angle, the radius of the inscribed circle, and the rectangle contained by the straight lines drawn from the centre of that circle to the angles at the base; to construct the triangle. Let the angle ABC be equal to the vertical angle, which bisect by the straight line BD. Let D (ii. 14.) be the centre of a circle which would touch BA and BC; B IN and let E and F be the points of con- perpendicular to AC. Then since BA and BC make equal angles with BG which passes through the centre, the arcs AD, DI, which they cut off, are equal; .. the angle ACD is equal to DCI, i. e. ACB is bisected by CD, or D is the centre of the circle inscribed in the triangle ABC; .:. K is the point of contact, and DK=DE. But (Eucl. vi. C.) the rectangle AD, DC is equal to the rectangle DK, DG, i. e, to the rectangle DE, DG, which is equal to the given rectangle. G (29.) Given the base, one of the angles at the base, ) and the point, in which the diameter of the circumscribing circle drawn from the vertex meets the base; to construct the triangle. Let AB be equal to the given base, and C the given point. Bisect AB in D, and draw DE at right angles to AB. Upon AC describe a segment of a circle containing an angle equal to twice the difference between the given angle and a right angle; and let it meet DE in 0. Join AO, OC; and produce CD B |