« ΠροηγούμενηΣυνέχεια »
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 A 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 ó lines will trisect the AABC.
Join AE, AF.
DEMONST.-1. Because AG is || to de, the A ADG = A AEG, upon the same base Ag, and between the same || 5. (Prop. XXXVII.)
2. To each add the A ABG; .. the trapezium ADGB = A AEB (Ax. 2).
3. In like manner, the trapezium ADHC = A AFC; .. by addition, ADGB + ADHC = AEB + AFC (Ax. 2).
4. Subtract each from the A ABC; .. the A GDH = A AEF (Ax. 3). 5. Now the Δ ΑEF
A AFC (Prop. XXXVIII.); .. the A 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 om, of which
the diam. is BC, E any pt. in Bc. Join AE, ED. The A ACE = DCE, and the A ABE = ADBE.
DEMONST.-1. Because AF = FD (Prop. EE), .. the A AFC
A DFC, and the A AFE = DFE (Prop. XXXVIII.); .., by addition, A ACE = A DCE (Ax. 2).
2. But the A ABC = A DBC (Prop. XXXIV.); .., by subtraction, A ABE = ADBE (Ax. 3).
Therefore, if from any pt. &c.-Q. E. D.
Gen. Enun.—To bisect a trapezium by a line drawn from one of its angles.
PART. Enun. and Const.-Let ABCD be the gn. trapezium, BAC the from which it is required to be bisected. Draw the diagonals AD, BC. Bisect bc, which is opposite the 2 BAC, in E. (Prop. X.)
Join AE, C ED; and through E draw FEG || to ad. (Prop. XXXI.) Join AHG. Then AHG bisects the trapezium.
DEMONST.-1. Because CE = EB, ... the A AEC AEB, and the A DEC = BED (Prop. XXXVIII.) ; figure AEDC figure AEDB (Ax. 2).
2. Also the A AEG = A DEG, upon the same base EG, and between same || $ EG, AD (Prop. XXXVII.). Take away the common part Eng; .. the À AEH = A DHG (Ax. 3).
3. Hence the A ACG = = figure AEDC – A DHG = figure AEDC AAEH = figure AEDB Δ ΑΕΗ trapezium AGDB.
Wherefore the trapezium ABCD has been bisected by a st. line AG drawn through one of the 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.
PART. ENUN. — Let the = AS ABC, DBC be upon
the same base BC, and upon
the of it; then they are between thesame ll S.
CONST. and DEMONST. (Ad absurd.) :
1. Join AD; and if ad be not ll to BC, through a draw AE || to BC (Prop. XXXI.), and join Ec; then the A
A EBC, upon the same base, and between the same || $. (Prop. XXXVII.)
2. But the A ABC = A DBC (Hyp.); .. the A DBC = A 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.
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,
Ac, in D and E. (Prop. IX.) Join
Then dE is ll to BC.
A CED; .. the ABDE =
Wherefore, if two sides &c.-Q. E. D.
PROP. XL. THEOR.
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 = A ABC, DEF be upon = bases BC, EF, in the st. line
and towards the same parts ; then they are between the same || S.
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,
= base EF, and between same || S. (Prop. XXXVIII.)
2. But the A ABC = A DEF (Hyp.); .. the A DEF
AGEF (Ax. 1), the > =“, which is impossible.
2. Hence Ag is not ll to BF; and, in like manner, no other line except ad is II to it.
Wherefore equal As &c.-Q. E. D.
A G D
PROP. UU. THEOR. 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.
Part. Enun.-1. Let the trapeziums ABCD, EBCF be upon the same base Bc, and between the same || 8 AF, BC; and let the side AD = side EF. Then the trapezium ABCD
A EBC, upon same base BC,
bases AD, EF, and between same || 8. (Prop. XXXVIII.); -., by addition, trapezium ABCD = trapezium EBCF.
2. Conversely, let the trapeziums be =, and between the same || $; then the side AD = the side EF.
DEMONST. (Ad abs.)
1. For, since trapezium ABCD trapezium EBCF, and, as before, the A DBC = A EBC, .., by subtraction, A 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 ABDG = A CEF, since they are upon = bases GD, EF, and between same || 8. (Prop. XXXVIII.)
4. But the A ABD = A CEF; .. ABDG = A ABD, the <= >,
which impossible. 5. .. AD is not unequal to EF, i. e. it is to it. Wherefore trapeziums &c.-Q. E. D.