(8.) Given one angle, a side opposite to it, and the sum of the other two sides ; to construct the triangle. a A B Let AB be the given side. Upon it describe a segment of a circle containing an angle equal to half the given angle ; and from A draw AC equal to the given sum of the two sides ; join BC; and make the angle CBD equal to BCD; ABD is the triangle required. Since the angle DCB is equal to DBC, DB is equal to DC; .:. AD, DB together are equal to the given sum. And the angle ADB is equal to DBC, DCB, i. e. to twice DCB, and .. is equal to the given angle. E а B (9.) Given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with the base ; to construct the triangle. On any line AB describe a segment of a circle containing an angle equal to the given angle. Bisect · AB in C; and at C make the angle BCD equal to the given angle which the bisecting line makes with the base; produce it, if necessary, till CE is equal to the given line; draw EF, EG respectively parallel to DA, DB; EFG is the triangle required. For FE and EG being respectively parallel to DA, DB, FC : CE :: (AC=) CB : CD :: CG : CE, ... FC=CG, and EC, which is equal to the given line, bisects the base; and the angles FEC, CEG being equal to the angles ADC, CDB, FEG is equal to ADB, i. e. to the given angle. (10.) Given the vertical angle, the perpendicular drawn from it to the base, and the ratio of the segments of the base made by it; to construct the triangle. A B Take any line AB, and on it describe a segment of a circle containing an angle equal to the given angle. Divide AB in C, in the given ratio; and from C draw the perpendicular CD, from which cut off DE equal to 'the given perpendicular. Join DA, DB; and through E draw FEG parallel to AB; DFG is the triangle required. F E G Since FG is parallel to AB, FE EG AC: CB, i. e. in the given ratio, and DE is equal to the given perpendicular, and FDG to the given angle. (11.) Given the vertical angle, the base, and a line drawn from either of the angles at the base to cut the opposite side in a given ratio; to construct the triangle. Let AB be equal to the given base; and divide it at D in the given ratio. On AD describe a segment of a circle containing an angle equal to the given angle; and from B draw BE equal to the given line. Join AE, ED; and from B draw BC parallel to DE, and meeting AE produced in C; ABC is the triangle required. E D B For AB is the given base ; BE is the given line; and AE : EC :: AD : DB, i.e. in the given ratio; and the angle at C is equal to AED, i. e. to the given angle. K ERC L B G (12.) Given the perpendicular, the line bisecting the vertical angle, and the line bisecting the base; to construct the triangle. From any point C in the indefinite line AB, draw a perpendicular CD equal to the given perpendicular; and with D as centre, and radii equal to the two given lines describe circles cutting AB in E and F. Through E draw GEH perpendicular to AB; join DE, DF; and produce DF to meet HE in G. Bisect DG in 1; and draw 10 at right angles to DG, meeting GH in 0. With the centre 0, and radius OG describe a circle cutting AB in K and L; join DK, DL; DKL is the triangle required. Join OD, OK, OL. Since Ol bisects DG at right angles, OD=OG, and the circle passes through D. And since OE is perpendicular to KL, KE=EL, or KL is bisected by DE, which is equal to the given bisecting line; and the arc KG=GL, and the angle KDF is equal to FDL, or the angle KDL is bisected by DF, which is equal to the given line; and DC was made equal to the given perpendicular. = (13.) Given the line bisecting the vertical angle, PP the line bisecting the base, and the difference of the angles at the base ; to construct the triangle. Construct a right-angled triangle FDC, (see last Fig.) having its hypothenuse FD equal to the given line which bisects the vertical angle, and the angle FDC equal to half the given difference of the angles at the base. Produce FC both ways; and to it from D draw DE equal to the given line which bisects the base. Draw HEG parallel to DC, meeting DF produced in G. Bisect GD in I; and from I draw 10 at right angles to DG, meeting GH in O. With the centre 0, and radius OG, describe a circle, cutting FC produced in K and L; join DK, DL; DKL is the triangle required. For KE=EL, i, e. the base KL is bisected by DE, which is equal the given line; and the angle KDF is equal to FDL, being on equal circumferences KD, DL; i. e. the vertical angle KDL is bisected by DF, which was made equal to the given bisecting line. Also (iii. 5.) the difference between the angles DLK and DKL is equal to twice the angle FDC, i. e. to the given angle. a (14.) Given the vertical angle, and the line drawn to the base bisecting the angle and the difference between the base and the sum of the sides ; to construct the triangle. Let ABC be equal to the given angle, and BD the line bisecting it. Make BE and BF, each equal to half the given difference. From F draw FO perpendicular to BC, meet B A D G ing BD in 0. With the centre 0, and radius OF describe a circle, it will touch AB in E (Eucl. iv. 4.). Through D draw a line AC touching the circle in G; ABC will be the triangle required. For (Eucl. iii. 36. Cor.) AE=AG, and GC=CF, :: AC is equal to AE and CF together; whence the difference between AC and the sum of the sides AB, BC is equal to BE, BF together, i.e. to the given difference. Also BD is equal to the given line, and it bisects the angle ABC, which is equal to the given ver tical angle. (15.) Given the line bisecting the vertical angle, the perpendicular drawn to it from one of the angles at the base, and the other angle at the base ; to construct the triangle. G F D E K B Let AB be equal to the given bisecting line; and upon it describe a segment of a a circle containing an anglè equal to the given angle. Draw BC perpendicular to AB, and make BD equal to the given perpendicular. Bisect AB in E; join ED, and produce it to F; join FA, FB; and through D, draw GDH ; parallel to AB. In FB produced take BI equal to BH. Join AI; AFI is the triangle required. Join IG, cutting AB in K. Because GH is parallel to AB, and FE bisects AB, it also bisects GH, i.e. GD = DH; but HB also is equal to Bl; .. BD is parallel to GI, and IK is half of IG, and ;. equal to BD the |