fall in together with the Sides that are equal to them, AB, AC, and this in fuch fort (c) that the three Points (c) Per (L, F, I) fhall fall upon the three Points, (A, B, C). Axio, 8. Therefore the whole Bafe FI will alfo fall upon the whole Bafe B C. But then the Angles F, B, and likewife those I, C, and the whole Triangles will mutually (congruere) agree to each other. All therefore by Axiom 7th are equal. 2. E.D. Which was the Thing to be demonftrated.

Coroll. (1.) Hence we may also in another way mea-Fig. 78. Sure the Line AB, altho otherwise impracticable by reafon of fome Obftacle, as a River, &c. between the Extremities thereof. For from any Point whatsoever, as the Point C, let the Angle ACB be observed, and then let the Lines AC, BC be measured and in any acceffible Plane let there be measured about the Angle F, which is equal to the Angle C, two Lines FD and FE, which are equal to the Lines AC and BC respectively. And then there will be the acceffible Line DE equal to the inacceffible AB. Q.E.I.

Coroll. (2.) Hence alfo, those who play at Billiards Fig. 79. with Ivory Balls may learn how by the Reflexion of their own to hit and remove their Adverfaries Ball. For let B be the Ball to be striken, A that which is to Strike it, and CD the Rectilinear Plain. Let the Line BE be perpendicular to the Line CD, and DE be equal to DB. If the Ball A be stricken and carried along the right Side AFE unto the Point F, it will there be fo reflected that after the Reflexion it will tend unto B. For in the Triangles BFD, EFD, the Side FD is common to both, and the Side DB is equal to the Side DE; and the Angles at D are equal, as being right ones. The whole Triangles therefore are equal: and therefore the Angle BFD, which is equal to the Angle DFE, is equal to AFC, the Angle AFC be-* Per 15.1.1. ing vertically oppofite to DFE. Wherefore, feeing the Angle AFC is the Angle of Incidence, which in fuch cafes is equal to the Angle of Reflexion, it is manifeft that BFD, which hath been proved equal to AFC, is the Angle of the Reflexion of the Ball A, and that the Ball tending towards E is in the Point F fo reflected as to hit the Ball B. QE, D.

*

Scholium

Fig.25.

(2) Per

Axi. 8. (b) Per Axi. 8.

Fig.26.

Scholium or Obfervation.

BY
Y much the fame way of Reasoning whereby this
4th Propofition has been demonftrated, the follow-
ing Theorem, which we fhall have occafion to ufe by
and by, may be demonftrated alfo.

If in Two Triangles X, Z, the Sides BC and FI fhall be equal, and the Angles adjacent to thefe Two Sides equal alfo, viz. B and C equal to F and I; all the other Things, and the whole Triangles themfelves will be equal. For the Side FI laid upon the Side B C will agree, or thorowly concide with it (a). And then because the Angles B and C are equal to thofe F and I, when the Side FI is laid upon the Side BC: FL (b) will fall exactly upon BA, and IL upon CA. Therefore the Point L will fall upon the Point A (for if it fall without A, the Sides FL, IL would not fall upon the Sides BA, CA). Therefore all Things are equal by Axiom 7th.

PRO P. V. Theorem.

N an Ifofceles or Equicrural Triangle, the Angles at the Bafe (A, C) are equal.

Let the Triangle ABC be understood to be twice put, but in an inverted Pofture cha. Because therefore in the Two Triangles ABC, cba, the Side A B is by the Suppofition equal to the Side cb, and the Side CB to the 'Side a b, and the Angle B to the Angle b; the Angle A (c) Per 4.1.1. alfo at the Bafe will (c) be equal to the Angle c. Q. E. D. For as for the Angles C and c, they are the fame.

Fig. 26.

ΤΗ

IF

Corollary.

HEREFORE an Equilateral Triangle is alfo
Equiangular.

PRO P. VI. Theorem.

F in a Triangle (ABC) Two Angles (A and C) be equal, the Sides (AB, BC) which are oppoJite to thofe Angles are equal also.

Let

Let the Triangle ABC be fuppofed to be twice put, but in an inverfe fituation, cba; because therefore in the Triangles ABC, cba, one Side A'C is equal to one Side (ca) and the Angle A is equal to the Angle c, and the Angle C equal to the Angle a, all the other Things fhall be likewife (a) equal, and confequently A B fhall (a) Per Schol. be equal to the Side cb. Q. E. D. For as for the Lines Prop. 4. CB and cb they are the fame.

Coroll.

THEREFORE an Equiangled Triangle, is also
Equilateral.

Coroll. (2.) Hence, by the means of the Shadow of the Fig. 80. Sun, we may measure the Height of a Tower, or any elevated Point. For when the Sun is elevated 45 Degrees above the Horizon, the Shadow which the Tower cafts towards the Horizon will be exactly equal to its Height. For, by reafon that the Angle ACB is half a right Angle, the Angle BAC alfo * will be half a right one; and fo, by the force of the prefent Propofition, the Line AB11. Prop. 32. will be equal to the Line BC. The Line BC therefore being found by measuring, there is found at the Same time the Line AB, the Height of the Tower above the Horizon.

Coroll. (3.) The fame Thing alfo may be found without the Sun by the means of an Aftronomical Quadrant. For where the Angle of Elevation is half-right, there the Height of the Tower above the Obferver's Eye is equal to the distance of the fame Eye, from that Part of the Tower which is oppofite to it. The Distance therefore of the Eye from the Tower being given by meafuring, there is given at the fame time the Height of the Tower, Q. E, I.

The VIIth Propofition in Euclid is for the fake of the VIIIth, which without it will here be demonftrated.

PRO P. VIII. Theorem.

I among themselves respectively and their sides

*Per Corol.

1.

F Two Triangles (X,Z) have all their Sides equal Fig. 27. (AC equal to EF; CB to FI; AB to EI;) they will alfo bave all the Angles which are oppofite to equal Sides, equal: (C equal to F; A to E; B to I.)

For

Fig. 81.

Fig. 29.

For fuppofe the Side A B laid upon its Equal EI, if then the Point C falls upon F, the Triangles will in the Whole agree or coincide, and confequently all the Angles will be equal. But the Point C will fall upon the Point F. For,

From the Centre A let a Circle be defcribed with the Semidiameter EF; and from the Centre I let another Circle be defcribed with the Semidiameter IF; the Point C by reafon of the Equality of the Sides of both Triangles, will be in the Circumference of both Circles, and confequently in the Point E, the common InterSection of both thefe Circumferences. Q. E. D.

Bifect or Divide into two equal Parts a given right-lin'd Angle, as IAL.

From the Sides of the Angle take with a Pair of Compaffes two equal Lines, A B, A C; then from the Centres B and C defcribe two equal Circles cutting one another in F; which done draw the Line FA. This bifects the Angle.

For draw the Lines BF, CF; the Triangles FA B, FAC are to each other Equilateral; for the Sides AB, AC are by the Conftruction equal, as in like manner are the Sides BF, CF, they being Semidiameters of equal Circles; and AF is common to both Triangles. There(d) Per 8.1.1. fore the Angles B A F, CAF (d) are equal. Therefore the given Angle IAL is bifected. Q.E. F

Fig. 30.

Corollary.

HENCE we learn how an Angle may be divided into 4, 8, 16, &c. equal Angles, viz. by bifecting each Part again.

No

Scholium,

one hath hitherto taught the way of dividing Angles into all equal Parts whatsoever with a Pair of Compaffes, and a Rule.

Yet may you divide any given Angle mechanically into any equal Parts whatfoever, if from the Top of the Angle as

the

the Centre you defcribe an Arch between the Legs of the Angle, and divide that Arch into as many equal Parts as you require; for right Lines let down from A thro' the Points of the Divifion, will cut the Angle into fo many equal Parts.

PROP. X. Problem.

T. bife&t a finite given Line ( A B.)

13

Fig. 31.

Upon the given A B make an Equilateral (a) Trian- (a)Per.1.1.1. gle AGB. Bifect its Angle G (b) with the right Line (b) Per praGC. The fame fhall bifect the given Line A B.

ced.

For in the Triangles X, Z, the Side CG is common; and by the Construction G B, GA are equal, and the Angles contained between them AGC, BGC, are likewife equal. Therefore the Bafes AC, BC (c) (c) Per 4.1.1. are equal. The given Line therefore A B is bifected. Q. E. F.

But for Practice it is fufficient from the Centers A and B to defcribe two equal Circles, cutting one another in G and L, and fo to draw the right Line GL.

PROP. XI. Problem.

ROM a given Point (A) in a given right Line Fig. 32.
(LI) to raise a Perpendicular.

FRO

With a Pair of Compaffes take the equal Lines AC, AF. From the Centre C and F defcribe two Circles, cutting one another in B. The Line which is drawn from B to A will be the Perpendicular required.

For let the right Lines CB, FB be drawn. The Triangles X and Z are equilateral to one another. Therefore the Angles CAB, FAB are equal (a.) (a) Per 8.1.1. Therefore BA is (b) perpendicular to the Line (LI.)) Per def. 2. E. F.

In Practice this and the next are eafily performed by the help of a Square.

14.

· PROP.