PROP. QQ. PROB. GEN. ENUN.-To trisect a given triangle from a given point within it. PART. ENUN.-Let ABC be the gn. A, D the gn. pt. within it. It is required to trisect ▲ ABC by lines drawn from the pt. D. CONST.-Trisect the side BC in E and F (Prop. Z). Join DE, DF; and from a draw AG, AH respectively to them. (Prop. XXXI.) Join DG, DH, AD; then these three lines will trisect the ▲ ABC. Join AE, AF. DEMONST.-1. Because AG is | to DE, the ▲ ADGA AEG, upon the same base AG, and between the same || $. (Prop. XXXVII.) 2. To each add the ▲ ABG; .. the trapezium ADGB = ▲ AEB (AX. 2). 3. In like manner, the trapezium ADHC = ▲ AFC; .. by addition, ADGB + ADHC = AEB + AFC (Ax. 2). 4. Subtract each from the ▲ ABC; .. the ▲ GDH = Δ AEF (AX. 3). = A AEB = 5. Now the ▲ AEF A AFC (Prop. XXXVIII.) ; .. the ▲ GDH = trapezium ADGB = trapezium ADHC (Ax. 1). Wherefore the A ABD has been trisected from a gn. pt. D within it.-Q. E. F. PROP. RR. THEOR. GEN. ENUN.-If from any point in the diameter of a parallelogram straight lines be drawn to the opposite angles, they will cut off equal triangles. PART. ENUN. and CONST.-Let ABCD be a ", of which DFE (Prop. XXXVIII.) ; .., by addition, ▲ ACE = 2. But the ▲ ABC = ▲ DBC (Prop. XXXIV.); . subtraction, ▲ ABE = ▲ DBE (Ax. 3). Therefore, if from any pt. &c.-Q. E. D. by PROP. SS. PROB. GEN. ENUN.-To bisect a trapezium by a line drawn from one of its angles. PART. ENUN. and CONST.-Let ABCD be the gn. trape zium, BAC the from which it is required to (Prop. X.) Join AE, H B D Join AHG. Then AHG = DEMONST.-1. Because CE = EB,.. the ▲ AEC AEB, and the ▲ DEC = BED (Prop. XXXVIII.); .. the figure AEDC = figure AEDB (Ax. 2). 2. Also the ▲ AEG = ▲ DEG, upon the same base EG, and between same || $ EG, AD (Prop. XXXVII.). Take away the common part EHG; .. the ▲ AEHA DHG (AX. 3). 3. Hence the ▲ ACG = figure AEDC A DHG = figure figure AEDB Δ ΑΕΗ = AEDC Δ ΑΕΗ = trapezium Wherefore the trapezium ABCD has been bisected by a st. line AG drawn through one of the 4 BAC.-Q. E. F. PROP. XXXIX. THEOR. PART. ENUN.-Equal triangles upon the same base, and upon the same side of it, are between the same parallels. CONST. and DEMONST. (Ad absurd.): 1. Join AD; and if AD be not || to BC, through a draw AE || to BC (Prop. XXXI.), and join Ec; then the ▲ ABC = ▲ EBC, upon the same base, and between the same || s. (Prop. XXXVII.) 2. But the ▲ ABC = ▲ DBC (Hyp.); the ▲ DBC = ▲ EBC, the > = the <, which is impossible. 3. ... AE is not || to BC; and, in like manner, no other line except AD is to it. Wherefore =AS upon &c.-Q. E. D. Hence the following deduction : PROP. TT. THEOR. GEN. ENUN.-If two sides of a triangle be bisected, the straight line which joins the points of bisection is parallel to the third side. PART. ENUN.-Let ABC be any A; bisect the sides AB, A D GEN. ENUN.-Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. PART. ENUN.-Let the = ▲ ABC, DEF be upon = bases BC, EF, in the same st. line BF, and towards the same parts; then they are between the same || s. B E F CONST. and DEMONST. (Ad absurd.) :1. Join AD; and if it be not to BF, through a draw AG || to BF (Prop. XXXI.), and join GF; then the ABC =A GEF, upon the base EF, and between same || S. (Prop. XXXVIII.) .'. 2. But the ▲ ABC = ▲ DEF (Hyp.) the A DEF = Δ GEF (Ax. 1), the > = <, which is impossible. 2. Hence AG is not manner, no other line except AD is Wherefore equal $ &c.-Q. E. D. To this may be subjoined the following: : to BF; and, in like PROP. UU. THEOR. to it. GEN. ENUN.-Trapeziums upon the same base, having the sides opposite to the base equal, and lying between the same parallels, are equal to one another; and equal trapeziums upon the same base, and between the same parallels, have their sides opposite to the base equal. 8 PART. ENUN.-1. Let the trapeziums ABCD, EBCF be upon the same base BC, and between the same || AF, BC; and let the side AD side EF. Then the trapezium ABCD = = = = ▲ EBC, upon same base BC, and between the same (Prop. XXXVII.); and the ▲ ADB = A FEC, upon the bases AD, EF, and between same || 5. (Prop. XXXVIII.); .'., by addition, trapezium ABCD = trapezium EBCF. 2. Conversely, let the trapeziums be same; then the side AD = the side EF. DEMONST. (Ad abs.) and between the 1. For, since trapezium ABCD = trapezium EBCF, and, as before, the ▲ DBC = ▲ EBC, ..., by subtraction, ▲ ADB = A CEF. 2. Now, if AD be not = EF, let AD be the >; cut of GD = EF (Prop. III.); and join GB. 3. Then the ▲ BDG = A CEF, since they are upon = bases GD, EF, and between same s. (Prop. XXXVIII.) 4. But the ▲ ABD = ▲ CEF; .. ▲ BDG = ▲ ABD, the <= >, which is impossible. 5. .. AD is not unequal to EF, i. e. it is = to it. Wherefore trapeziums &c.-Q. E. D. |