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the gn. pt. All to the gn. st. line BC.Q. E. F.

It is clear that the pt. A cannot be so gn. that the line AD shall be in the same st. line with Bc.

The following Theorem may now be solved :

PROP. W. THEOR.

A

E

Gen. Enun.–Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that of which the base is bisected in the given point.

Part. Enun.- Let Bac be the vertical Z of any number of A$, whose bases pass through a gn. pt. D. Let the base Bc of the ABC be bisected in d; also let AEF be any other A, having the same vertical L, and whose base passes through D. The A ABC is < than the A AEF.

DEMONST, -1. Through c draw cg || to AB (Prop. XXXI.); then the L EBD = alternate z DCG (Prop. XXIX.); the verti. cal LEDB = vertical CDG (Prop. XV.); and the side BD = DC (Hyp.); .. = À GCD (Prop. XXVI.), and < than A CDF.

2. To each add the trapezium ACDE; .. the A ABC is < than A AEF.

3. The same may be proved with respect to any other A with the same vertical Z, whose base passes through D.

Wherefore, of all triangles &c.-Q. E. D.

B

F

EBD

PROP. XXXII. THEOR.

GEN. ENUN.-If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three inte

E

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B

с

D

rior angles of every triangle are equal to two right angles. PART. ENUN.—Let ABC be a A,

and let one of its sides bc be produced to D; then the exterior < ACD = the interior and opposite ZS CAB + ABC, and the three interior Zs

BC + BC + CAB = two rt. Zs.

Const.—Through the pt. c draw CE || to Ba.

DEMONST.–1. Because AB is || to CE, and AC meets them, .. the < BAC = alternate 2 ACE. (Prop. XXIX.)

2. Again, because AB is || to ce, and BD meets them, .. the exterior z ECD = interior

ABC. (Prop. XXIX.) 3. ... the 2 ACD = ZS ACE + ECD = ZS BAC + ABC.

4. Add to each the < ACB; .. the ZS ACD + ACB = { $ ABC + BCA + CAB (Ax. 2): but the < $ ACD + ACB = two rt. Zs (Prop. XIII.); .. the < $ ABC + BCA + CAB = two rt. $ (Ax. 1).

Wherefore, if a side of a triangle be produced, &c.-Q. E. D.

Cor. 1.—All the interior <5 of any rectilineal figúre + four rt. 28 = twice as many rt. Z $ as the figure has sides.

DEMONST.-1. For any rectilineal figure ABCDE may

be divided into as many as as the figure has sides, by drawing lines from each of its < s to any pt. F within it.

D

E

с

F

B

А

2. Now all the Zs of these As = twice as many rt. Z $ as there are As, i. e. as there are sides to the figure.

3. And these same zs= the <s of the figure + 2S at the pt. F, which is the common vertex of all the As = Zs of the figure + four rt. 2 (2 Cor., Prop. XV.); -. all the 2$ of the figure + four rt. 28 = twice as many rt. 2$ as the figure has sides.-Q. E. D. Cor. 2.-All the exterior Zs of

any

recti. lineal figure are together = to four rt. 29. DEMONST.-Because every

interior 2 ABC + its exterior _ ABD = two rt. Zs (Prop. XIII.); .. all the interior 2s + all the exterior ZS = twice as many rt.

Z as the figure has sides : i.e. = all the interior Zs + four rt. Zs (Cor. 1); the exterior ZS = four rt. Zs. -Q.E.D.

From this Theorem, which is an amplification of the 16th and 17th Propositions, it is clearly deducible that if one of the Ls of a À be a rt. , the sum of the other two is to a rt. 1 ; that in an isosceles rt. 41 A, each of the <s at the base is half a rt. (Prop. V.); that each of the <s of an equilateral A is one-third of two rt. ZS, or twothirds of a rt. L ; and that if two of the angles of any A are = to two of the angles of another, the third 2 of the one is to the third of the other. The converse of the Proposition may be thus partially enunciated and proved: —

A

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PROP. X. THEOR.

A

Gen. Enun.—If a straight line be drawn from one of the angles of a triangle, making the exterior = to the two interior and opposite angles, it shall be in the same straight line with the side of the triangle.

Part. Enun.-From the angular pt. c of the A ABC, let a st. line cd be drawn, so that the exterior . ACD the two interior and opposite 28 ABC + BAC; then CD is in the same st. line with Bc.

DEMONST.-Since the L ACD = Ls ABC + BAC, add to each side the L ACB; .. the LS ACD + ACB LS ABC + BAC + ACB ; but ZS ABC + ACB

two rt. 48 (Prop. XXXII.); .. LS ACD + ACB = two rt. 28 (Ax. 1); .. cd is in the same st. line with Bc. (Prop. XIV.)

Wherefore, if a st. line &c.—Q. E. D.
To this may be added the following deductions:-

BAC +

B

PROP. Y. PROB.

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GEN. ENUN.--To trisect a right angle; that is, to divide it into three equal parts.

Part. ENUN.–Let ABC be a right Z; then it is required to divide it into 3 = parts.

Const.-In sc take any pt. C, and upon bc describe an equilateral A BCD (Prop. I.); bisect the | DBC by the st. line BE (Prop. 9); then the Z S ABD, DBE, EBC are all equal.

DEMONST. - For the LDBC B (being one of the 28 of an equilateral A ) = įrds of a rt. Z; .. each of the 19 DBE, EBC (by construction)

grd of a rt. L; .. also, the 2 ABD = žrd of a rt. L; . . ABD= LDBE =

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L EBC.

H

Wherefore the rt. ABC has been trisected by the st. lines BD, BE.-Q. E. F.

PROP. Z. PROB.

C

Gen. EnUN.--To trisect a given finite straight line; that is, to divide it into three equal parts.

Part. Enun.—Let AB be the gn. st. line; then it is required to divide it into three parts.

Const.-Upon AB describe an equilateral A ABC (Prop. I.); bisect the / CAB, CBA by the st. lines AD, BD, meeting in D (Prop. IX.); through o draw DE || to AC, and DF || to BC (Prop. XXXI.); then AB will be trisected in the pts. E, F.

DEMONST.-1. Because ED is Il to Ac, the EDA = alternate

DAC (Prop. XXIX.) = = L DAE (Const.); .. AE = ED. (Prop. VI.)

2. In like manner, BF = FD.

3. Again, because de is || to ca, and DF to CB, . . exterior [ DEF = interior / CAB, and exterior / DFE = interior / CBA (Prop. XXIX.); .. the third L EDF of A

third / ACB of A ABC. (Prop. XXXII.) 4. Hence the A DEF is equiangular, and .. equilateral (Prop. VI. Cor.); and,. . AE (= DE) = EF (= FD) = FB.

Therefore the gn. st. line AB has been trisected in the pts. E and F.-Q. E. F.

A

E

B

DEF =

PROP. AA. THEOR.

Gen. ENUN.-If the three angles of a triangle be bisected, and one of the bisecting lines be produced to the opposite side, the angle contained by this line produced, and one of the others, is equal to the angle contained by the third, and a perpendicular drawn from the common point of intersection of the three lines to the aforesaid side.

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