the gn. pt. A to the gn. st. line вc.— 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.

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

of As, 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, and whose base passes through D. The A ABC is than the ▲ Aef.

DEMONST.-1. Through c draw CG to AB (Prop. XXXI.); then the EBD = alternate DCG (Prop. XXIX.); the vertical EDB = vertical

CDG

E

of any number

A

C

B

D

F

(Prop. XV.); and the side BD = DC (Hyp.); .'. ▲ ebd A GCD (Prop. XXVI.), and < than ▲ CDF.

=

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

3. The same may be proved with respect to any other ▲ with the same vertical Z, whose base passes through D. Wherefore, of all triangles &c.-Q. E. D.

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

rior angles of every triangle are equal to two right angles.

PART. ENUN.-Let ABC be a ▲, and let one of its sides BC be produced to D; then the exterior ▲ ACD = the interior and opposite <S CAB + ABC, and the three interior <s ABC BCA + CAB =

two rt. s.

CONST.-Through the

pt. c draw CE to BA.

E

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

2. Again, because AB is

meets them, .. the exterior ABC. (Prop. XXIX.)

3... the

BAC ABC.

to CE, and BD

ECD = interior

ACDS ACE + ECD = s

4. Add to each the

ACB; .. the S ACD + ACB = <$ ABC + BCA + CAB (AX. 2): but the SACD + ACB = two rt. 4s (Prop. XIII.); .. the ≤ ABC + BCA + CAB = two rt. s (Ax. 1).

S

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

COR. 1.-All the interiors of any rectilineal figure four rt. s = twice as many rt. s as the figure has sides.

DEMONST.-1. For any rectilineal figure ABCDE may be divided into as many

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

2. Now all the s of these s twice as many rt. s as there are ▲s, i. e. as there are sides to the figure.

E

3. And these same s = thes of the figure + <s at the pt. F, which is the common vertex of all the As = A

S

F

D

C

B

<s of the figure + four rt. 4s (2 Cor., Prop. XV.); .. all the s of the figure + four rt. <s twice as many rt. Zs as the figure has sides.-Q. E. D.

S

COR. 2.-All the exterior s of any rectilineal figure are together = to four rt. Zs. DEMONST. Because every interior ABC its exterior ABD = two rt. 4s (Prop. XIII.); .. all the interior s + all the exte

rior < S = twice as many rt. Zs as the figure has sides: all the interior

i.e.

S

s +

four rt. s (Cor. 1); .. all the exteriors four rt. s. -Q. E. D.

=

D

From this Theorem, which is an amplification of the 16th and 17th Propositions, it is clearly deducible that if one of the s of a ▲ be a rt. 2, the sum of the other two is to a rt. ; that in an isosceles rt. d A, each of the 4s at the base is half a rt. (Prop. V.); that each of the 4s of an equilateral ▲ is one-third of two rt. 4o, or twothirds of a rt. ; and that if two of the angles of any ▲ are to two of the angles of another, the third of the other. The converse of thus partially enunciated and

=

the one is to the third the Proposition may be proved:

of

PROP. X. THEOR.

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 ▲ ABC, let a st. line CD be drawn, so that the exterior / ACD = the two interior and opposite 4s ABC + BAC; then CD is in the same st. line with BC.

DEMONST.-Since the ACD =

ABC + BAC+ ACB; but $ ABC +

BAC + ACB =

two rt. 4s (Prop. B XXXII.); .'. ≤s ACD + ACB = two

A

rt. s (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:

PROP. Y. PROB.

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

PART. ENUN.-Let ABC be a right ; then it is required

to divide it into 3 =

parts. CONST.-In вс take any pt. c, and upon вc describe an equilateral ▲ BCD (Prop. I.); bisect the / DBC by the st. line BE (Prop. 9); then the ABD, DBE, EBC are all equal.

DEMONST. -For the DBC B (being one of the .. each of the rt. ; .. also, the ▲ ABD = LDBE = LEBC.

A

D

E

C

of an equilateral ▲) = rds of a rt. ≤ ; DBE, EBC (by construction) = 3rd of a rd of a rt. ;.. 2 ABD=

Н

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

PROP. Z. PROB.

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

=

C

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 ▲ ABC (Prop. I.); bisect the CAB, CBA by the st. lines AD, BD, meeting in D (Prop. IX.); through D draw DE Il to AC, and DF || to BC (Prop. XXXI.); then AB will be trisected in the pts. E, F.

DEMONST.-1. Because ED is A

I to AC, the EDA = alternate

E

B

F

Z DAC (Prop. XXIX.) = 4 DAE (Const.); .. AE = ed. (Prop. VI.)

FD.

2. In like manner, BF = 3. Again, because DE is terior DEF = interior interior CBA (Prop. XXIX.);

to CA, and DF to CB, .'. exCAB, and exterior / DFE = the third EDF of A

DEF= third ACB of ▲ ABC. (Prop. XXXII.)

4. Hence the ▲ 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.

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.