and opposite Angle

on the same fide; or makes the inward Angles AGH, CHG, on the fame fide equal to two right Angles; then are the right Lines (AB, CD) parallel.

Hyp. 1. Because by the Hypothefis the Ang. AGE = CHG, therefore the alternate Angles BGH, CHG are equal: and consequently ABa 15.1. and CD are parallel. b 27.1.

Hyp. 2. Because by the Hypothesis the Ang. AGH + CHG = two right Ang. = AGHC 13. 1. +BGH; therefore is d CHG = BGH: and d 3 ax. consequently AB, CD are e parallel. Q. E. D. e 27. 1.

ternate Angles DHG,
AGH, will be equal
to each other, and the
outward Angle BGE
will be equal to DHE,

the inward and opposite Angle on the fame fide; as alfo the inward Angles AGH, CHG on the Samè fide will be equal to two right Angles.

It is evident that AGH+CHG = two right Angles; for otherwise AB, CD would not be f parallel, which is contrary to the Hypothesis.f 13 ax. But also the Ang. DHG + CHG = twog 13. 1. right Angles: Therefore DHG = AGH h 3 ax. CO_1 15.1. BGE. Q.E. D.

b 3 ax.

For A+B= two

D

Angle, B must bebalso a right Angle. By the same Argument C and D are right Angles.

right Angles: therefore since A is a right

Lines any how ;

then because AB, EF are parallel, the Angle

C 29.1. AGI will be = EHI: also because CD and

EF are parallel, the Angle EHI will be

diax. DIG: Therefore d the Angle AGI = DIG.

e 27. 1. Whence AB and CD are parallel. Q.E.D.

a right Line AE, parallel to a right Line BC given.

From the Point

A draw a right Line AD to any given Point of the given right Line; with which at the Point

A

A thereof make a an Angle DAE = ADC: then a 27.1. will AE and BC be parallel. Q. E. D.

If one Side (BC) of any Triangle E (ABC) be continued out; the outward Angle ACD

D shall be equal to

the two inward opposite Angles A, B: and the three inward Angles A, B, ACB, of a Triangle, Shall be equal to two right Angles.

From C draw CE parallel to BA; then isa 31. 1. the Angle Ab = ACE, and the Angle Bbb 29. 1. ECD: Therefore A+B=ACE + ECD = c 2 ax. ACD. Q. E. D. d 19 ax.

Again, ACD + ACB = two right Angles: e 13.1. Therefore A+B+ACB = two right An-fi ax. gles. Q. E. D.

COROLLARIES.

1. The three Angles of any Triangle taken together, are equal to the three Angles of any other Triangle taken together. From whence it follows,

2. That if in one Triangle two Angles (taken severally or together) be equal to two Angles of another Triangle, (taken severally or together) then is the remaining Angle of the one, equal to the remaining Angle of the other. In like manner, if two Triangles have one Angle of the one, equal to one of the other; then is the Sum of the remaining Angles of the one Triangle, equal to the Sum of the remaining Angles of the other.

3. If one Angle in a Triangle be a right Angle, the other two are equal to a right one.

Like

Likewise that Angle in a Triangle, which is equal to the other two, is itself a right Angle.

4. In an Ifofceles Triangle, if the Angle made by the equal Sides is a right one, the other two upon the Base are each of them half a right Angle.

5. An Angle of an equilateral Triangle is two thirds of a right Angle: for of two right Angles is equal to of one.

SCHOLIU M.

By the help of this Proposition, you may know how many right Angles the inward and outward Angles of a right-lined Figure make; as will appear by these two following Theorems.

THEOR. I.

All the Angles of a right-lined Figure do together make twice as many right Angles, abating four, as there are Sides of the Figure.

From any Point within the Figure draw right Lines to all the Angles of the Figure, which shall resolve the Figure into as many Triangles as there are Sides of the Figure. Since therefore in every Triangle the Sum of all the Angles is two right Angles, all the Angles of the Triangles taken together will make up twice as many right Angles as there are Sides. But the Angles about the said Point within the Figure make four right Angles therefore if from the the Angles of all the Triangles, you take away the Angles which are about the faid Point, the remaining Angles, which make up the Angles of the Figure, will make twice as many right Angles, abating four, as there are Sides of the Figure. QE. D.

COROL..

Hence all right-lined Figures of the fame Species or Kind have the Sums of their Angles equal.

THEOR. II.

All the outward Angles of any right-lined Figure, taken together, make four right Angles.

For every inward Angle of a Figure, with the outward Angle of the same, make two right Angles; therefore all the inward Angles together, with all the outward Angles, make twice as many right Angles as there are Sides of the Figure. But (as it was just now shewn) all the inward Angles, together with four right Angles, make twice as many right Angles as there are Sides of the Figure; therefore the outward Angles are equal to four right Angles. Q.E. D.

COROL.

The Sum of the outward Angles of any rightlined Figure is equal to the Sum of the outward Angles of any other right-lined Figure.

PROP. XXXIII..

C

A