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PROP. XXVI. THEOR.

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GEN. ENUN.-If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each, then shall the other sides be equal, each to each, and the third angle of the one to the third angle of the other.

PART. ENUN.—Let ABC, DEF be two As, which have the < $ ABC, BCA = Z $ DEF, EFD, each to each ; viz. Z ABC

Z DEF, and BCA = 2 EFD; also let one side of a ABC=one side of A DEF, towards the same parts; then the other sides shall be each to each, and the third 2 BAC shall be = third 2 EDF.

Case 1.--Let the sides adjacent to the = <s in the two As be = to each other, i.e. let BC = EF; then, if al be not = DE, one of them must be > the other.

Const.—Let AB be the > of the two; make B GB = DE (Prop. III.), and join gc.

DEMONST. (Ad abs.)-1. Now because GB = DE, and BC = EF (Hyp.); .. in the A$ GBC, DEF, the two sides GB, BC = DE, EF, each to each, and the < GBC = _ DEF (Hyp.); .. the base GC = base df, and the A GBC = A DEF. (Prop. IV.) Also the remaining 2$ are =, each to

A

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G

CE

F

AB

A

D

each ; .. the 2 GCB = ZDFE = 2 ACB (Hyp.), i. e. the less = the greater, which is impossible.

2. .. AB is not unequal to DE, i.e. AB=DE.

3. Also BC = EF (Hyp.); .. the two AB, BC = DE, EF, each to each, and 2 ABC = DEF; .. the base AC = base DF, and the third Z BAC = third EDF. (Prop. IV.)

Case 2.—Let the sides opposite to = <3 in each a be i. e. let AB = DE. Then if BC be not = EF, let Bc be the greater of them.

Const.-Make BH = Er (Prop. III.), and join an DEMONST. -1. Now because BH = EF,

and DE (Hyp.); ... the two AB, BH = DE, EF, each to each, and the 2 ABH = Z DEF (Hyp.); .. the base AH = base DF, and the A ABH = A DEF, and the remaining <s are also respectively equal (Prop. IV.): .. the Z AHB = DFE = 2 ACB (Hyp.), i. e. the exterior < of A ACH = interior and opposite 2, which is impossible. (Prop. XVI.)

2. .. bc is not unequal to EF, i. e. BC=EF. 3. Also AB = DE (Hyp.);

the two AB, BC = DE, EF, each to each, and the 2 ABC = 2 DEF (Hyp.); .. the base Ac = and the third 2 BAC = the third 2 EDF. (Prop. IV.)

Wherefore, if two As have &c.—Q. E. D.

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CE

H

base DF,

G

This Theorem contains the third and last condition of the equality of two or more As, and, as in the Fourth and Eighth Propositions, it involves also the equality of the areas and the remaining sides and L. For since the two sides AB, BC = DE, EF, each to each, and contain ELS, not only does the base AC =

but the À ABC = DEF,

and the / BAC = | EDF, by Prop. IV. The following are useful deductions :

base EF,

A

AB

AC

B

D

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PROP. T. THEOR. GEN. ENUN.— If, in an equilateral or isosceles triangle, a straight line be drawn to the base, from the angle contained by the two equal sides, and bisecting the angle; it shall also bisect the base; or if it bisect the base, it shall also bisect the angle; and if it bisect either the angle or the base, it shall be perpendicular to the base.

Part. Enun.-Let ABC be an equilateral or isosceles A, having the side AB = AC. From the angular pt. A draw Ad to the base BC, and, first, let it bisect the L BAC (Prop. IX.); then BC shall also be bisected in D,

and ad shall be I to BC. DEMONST.

Because (Hyp.), and Ad is common to A8 ABD, ACD ; : . the two sides BA, AD = CA, AD, each to each, and the included BAD L CAD (Hyp.); .. the base bd = base cp (Prop. IV.); and the LBDA = LCDA, and they are adjacent 28; .. BC is bisected in D, and the line ad is I to BC (Def. 10).

Next let Bc be bisected in D; then also AD shall be I to BC, and bisect the L BAC.

DEMONST.-1. Because BD CD (Hyp.), and da is common to AS ADB, ADC; .. the two BD, DA = CD, DA, each to each ; and the base AB base Ac (Hyp.); . i the LBDA = L CDA, and they are adjacent /S; .. each of them is a rt. L.

2. Also because the side AB = AC; .'. the L ABC = ACB (Prop. V.); . ... the 4S ADB, ABD = L ADC, ACD, each to each ; and Ad is opposite to = ls in each; .. the third L BAD = CAD (Prop. XXVI.): i. e. the / BAC is bisected by AD, which has been shown to be 1 to bc.

Wherefore, if in an equilateral &c.-Q. E. D.

PROP. U. THEOR.

Gen. Enun.-If a straight line, drawn I to the base of a triangle from the opposite angle, bisect either that angle or the base, the sides containing the angle shall be equal.

Part. Enun.—Let ABC be a A. Through a draw AD I to bc (Prop. XII.), and, first, let Bc be bisected in D. Then the side AB shall be = to AC.

DEMONST.-Because BD= CD (Hyp.), and da is common; .., in AS ADB, ADC, the two BD, DA = CD, DA, each to each, and the rt. L

rt. / CDA (Def. 10); ... the base AB = base AC. B (Prop. IV.)

Again, let the L BAC be bisected by the I Then, because the 2 $ BAD, ADB = Z SCAD, ADC, each to each, and ad is opposite to = 48 in each ; •. also the base AB = base ac. (Prop. XXVI.)

Wherefore, if a st. line, &c.—Q. E. D.

BDA =

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AD.

PROP. V. PROB.

Gen. ENUN—Through a given point, to draw a straight line which shall make equal angles with two straight lines given in position.

PART. ENUN.-Let AB, CD be two st. lines gn. in position; E the gn.pt. Then

A А it is required

draw a line through E, which shall make = Ls with

B

G

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AB, CD.

E

D

с

Const.–Produce AB, co to meet in F (Post. 2); bisect the L AFC by the st. line rg (Prop. IX.); through draw Eg 1 to FG (Prop. XII.), and produce Eg both ways to B and D. Then BD is the line required.

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Because the LBFG = LDFG, and the at G are rt. 28 (Const.); .., in A ® BGF, DGF, the 2 BFG, FGB = DFG, FGD, each to each, and rg is opposite to the 28 in each ; ... the third / FBG = the third / FDG. (Prop. XXVI.)

Wherefore, through the gn. pt. E, a st. line BD has been drawn making Zs with the gn. st. lines AB, CD.Q. E. F.

PROP. XXVII. THEOR.

A.

E

B

H

F

Gen. Enun.—If a straight line, falling upon two other straight lines, makes the adjacent angles equal to one another, these two straight lines are parallel

PART. ENUN.—Let the st. line EF, falling upon the two st. lines AB, CD, make the alternate ZS AEF, EFD = to one another; then AB is Il to co.

Const. and DEMONST. -1. For, if not, AB and CD, being produced, will meet either towards B, D, or a, C. Let them be produced and meet towards B, D, in the pt. G; then the figure EFG is a s, of which the exterior Z AEF is > the interior opposite < EFG (Prop. XVI.); but these < s are also = (Hyp.), which is impossible; .. AB and cd, being produced, do not meet towards B, D.

2. In like manner it may be shown that they do not meet towards A, c.

3. But st. lines which never meet either

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