(6.) If from one of the equal angles of an isosceles triangle any line be drawn to the opposite side, and from the same point, a line be drawn to the opposite side produced, so that the part intercepted between them may be equal to the former; the angle contained by the side of the triangle and the first drawn line is double of the angle contained by the base and the latter. Let ABC be an isosceles triangle, having the side AB equal to AC. From B draw any line BD, and also BE cutting off DE equal to DB; the angle ABD is double of CBE. B E For the angle DCB is equal to the two DEB, CBE, i. e. to the two DBE, CBE, or to DBC and twice CBE; but DCB is equal to ABC, .. ABC is equal to DBC and twice CBE, and taking away the angle DBC, which is common to both, the angle ABD is equal to twice CBE. (7.) If from the extremity of the base of an isosceles triangle, a line equal to one of the sides be drawn to meet the opposite side; the angle formed by this line and the base produced, is equal to three times either of the equal angles of the triangle. Let ABC be an isosceles triangle having the side AB equal to AC. From C to AB (produced if necessary) draw CD equal to AC, and let BC be produced; the angle DCE is equal to three times the angle ABC. Since CA is equal to CD, the angle CAD is equal to CDA, .. CDA and twice ABC are together equal to two right angles, and .. are equal to CDA, CDB; whence CDB is double of ABC. Now (Eucl. i. 32.) the angle DCE is equal to the two angles CDB, CBD and consequently is equal to three times the angle ABC. (8.) The sum of the sides of an isosceles triangle is less than the sum of the sides of any other triangle on the same base and between the same parallels. Let ACB be an isosceles triangle, and ADB any other triangle on the same base, and between the same parallels AB, ED; AC and CB together will be less than AD and DB. E A B Since EC is parallel to AB, the angle ECA is equal to CAB; and for the same reason DCB is equal to CBA; but CAB being equal to CBA, ECA is equal to DCB; .. AC and BC drawn from two given points A and B on the same side of the line ECD given in position make equal angles with the line, .. (i. 6.) they are together less than any other two lines AD, DB, drawn from the same points to that line. (9.) If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side the part of it intercepted by a perpendicular from the vertex will have to one of the equal sides, the same ratio that the segment of the base has to the perpendicular upon the base. Let ABC be an isosceles triangle, having the side AB equal to AC. From Band A let fall perpendiculars BD, AE; then will BF: AC :: BE : EA. B D E C Since the angles BDA, AEC are right angles, and the angle DAF common to the two triangles FAD, EAC, .. the triangles are similar. But the triangle BFE is similar to AFD, and .. to EAC; whence BF: BE AC: AE, and BF: AC :: BE: EA. (10.) If from any point in the base of an isosceles triangle lines be drawn to the opposite sides, making equal angles with the base; the triangles formed by these lines, the segments of the base, and the lines joining the intersections of the sides and the angles opposite, will be equal. From any point D in AC the base of the isosceles triangle ABC let DE, DF be drawn making the angles CDE, ADF equal to one another; join AE, CF; the triangles AED, CDF are equal. E Since the angle ADF is equal to the angle EDC, and_FAD=ECD, the triangles ECD, angular, and AD: DC :: FD: DE. FAD are equi- angle FDA is equal to EDC, add to each the angle FDE, .. the angle ADE = CDF; hence the sides about the equal angles are reciprocally proportional, and .. (Eucl. vi. 15.) the triangles ADE, FCD are equal. N (11.) If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides; these together shall be equal to a perpendicular drawn from either extremity of the base to the opposite side. Let ABC be an isosceles triangle, from any point D in the base of which, let DE, DF be drawn perpendicular to the sides; and from Blet BG be drawn perpendicular to AC; BG is equal to DE and DF together. E B G P Since the angle EBD is equal to the angle at C, and the angles at E and F are right angles, the triangles BED, DFC are equiangular, and .. BD : DC :: DE : DF, comp. BC: DE+ DF :: DC: DF. But BG being parallel to DF, DC: DF :: BC: BG, whence BC: BG :: BC: DE+ DF, and .. BG-DE+ DF. (12.) Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in the given point. Let BAC be the vertical angle of any number of triangles, whose bases pass through a given point P, and let BC be bisected in P: ABC is less than any other triangle ADE. D From C draw CF parallel to AB; then the angle DBP is equal to PCF, and the vertically opposite angles DPB, CPF are equal, and BP = PC, ..the triangle DBP is equal to the triangle PCF, and .. DPB is less than CPE; add to each the trapezium ADPC, and ABC is less than ADE. In the same manner ABC may be proved to be less than any other triangle whose base passes through P. (13.) If from the angles at the base of a triangle perpendiculars be let fall on a line which bisects the vertical angle; the part of this line intercepted between these perpendiculars will be bisected by a perpendicular from the middle of the base. From A and B let perpendiculars AD, BG be drawn to the line CD which bisects the angle at C; the part GD will be bisected by a perpendicular EF from E the middle point of the base AB. H B G F E Produce BG, FE to H and I. Then IE being parallel to HB, and AE= EB, .. (Eucl. vi. 2.) AI= IH. Also since AD is parallel to HG and IF, DF FG:: AI: IH, whence DF FG, and DG is bisected in F. = (14.) If from one of the angles at the base of a triangle a line be drawn parallel to the opposite side, and from any point in it lines be drawn making any angles with the sides (produced, if necessary); they will have the same ratio that lines have, which are drawn parallel to them from the other angles, and terminated by the same sides. |