1

fall in together with the Sides that are equal to them, AB, AC, and this in such fort (c) that the three Points (c) Per (L, F, I) shall fall upon the three Points, (A, B, C). Axio. 8. Therefore the whole Base FI will also fall upon the whole Base B C. But then the Angles F, B, and likewise 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 demonstrated.

Coroll. (1.) Hence we may also in another way mea-Fig. 78. Jure the Line AB, altho otherwise impracticable by reaSon of Some Obstacle, 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 accessible 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 B C respectively. And then there will be the accessible Line DE equal to the inaccessible AB. Q.E.I.

Coroll. (2.) Hence also, those who play at Billiards Fig. 79. with Ivory Balls may learn how by the Reflexion of their own to hit and remove their Adversaries 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 so 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 Dare 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 opposite to DFE. Wherefore, seeing the Angle AFC is the Angle of Incidence, which in fuch cases is equal to the Angle of Reflexion, it is manifest 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 Eis in the Point F fo reflected as to hit the Ball B. Q. E. D.

Scholium

Fig. 25.

(a) Per

Axi. 8.

(b) Per Axi. 8.

Fig. 26.

Scholium or Obfervation.

BY much the fame way of Reasoning whereby this 4th Propofition has been demonstrated, the following Theorem, which we shall have occafion to use by and by, may be demonstrated also.

If in Two Triangles X, Z, the Sides BC and FI shall be equal, and the Angles adjacent to these Two Sides equal also, viz. B and C equal to F and I; all the other Things, and the whole Triangles themselves will be equal. For the Side FI laid upon the Side BC will agree, or thorowly concide with it (a). And then because the Angles B and C are equal to those F and I, when the Side FI is laid upon the Side BC: FL (1) will fall exactly upon B A, 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 BА, СА). Therefore all Things are equal by Axiom 7th.

PROP. V. Theorem.

Nan Isofceles or Equicrural Triangle, the Angles at the Base (A,C) are equal.

Let the Triangle ABC be understood to be twice put, but in an inverted Posture cha. Because therefore in the Two Triangles ABC, cba, the Side AB is by the Supposition equal to the Side cb, and the Side CB to the Side ab, and the Angle B to the Angle b; the Angle A (c) Per 4.1.1. also at the Base 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.

Corollary.

!

THEREFORE an Equilateral Triangle is also
Equiangular.

PROP. VI. Theorem.

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

Let

Let the Triangle ABC be supposed to be twice put, but in an inverse situation, 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 shall be likewise (a) equal, and consequently A B shall (a) Per Schol. be equal to the Side cb. 2. F.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 casts towards the Horizon will be exactly equal to its Height. For, by reason that the Angle ACB is half a right Angle, the Angle B AC also * will be half a right one; and fo, * Per Corol. by the force of the present Proposition, the Line AB 11. Prop. 32.

will be equal to the Line BС. 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 also may be found without the Sun by the means of an Astronomical Quadrant. For where the Angle of Elevation is half-right, there the Height of the Tower above the Observer's Eye is equal to the distance of the same Eye, from that Part of the Tower which is opposite to it. The Distance therefore of the Eye from the Tower being given by measuring, there is given at the same 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.

PROP. VIII. Theorem.

1. 1.

F Two Triangles (X,Z) have all their Sides equal Fig. 27. EF;

Iamong it themjerves respectively aberasad to

CB to FI; AB to EI;) they will also have all the Angles which are opposite to equal Sides, equal: (C equal to F; A to E; B to I.)

For

3

Fig. 81.

Fig. 29.

For suppose 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 consequently 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 described with the Semidiameter EF; and from the Centre I let another Circle be described with the Semidiameter IF; the Point C by reason 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 these Circumferences. Q. E. D.

PROP. IX. Problem.

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

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

For draw the Lines BF, CF; the Triangles FAB, FAC are to each other Equilateral; for the Sides AB, AC are by the Construction 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 AF, CAF (d) are equal. Therefore the given Angle I AL is bisected. Q.E. F.

Corollary.

HENCE we learn how an Angle may be divided into 4, 8, 16, 5c, equal Angles, viz, by bisecting each Part again.

Scholium,

No one hath hitherto taught the way of dividing An⚫gles into all equal Parts whatsoever with a Pair of

Compasses, and a Rule.

Fig. 30.

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

the

the Centre you describe 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 Division, will cut the Angle into so many equal Parts.

T

PROP. X. Problem.

O bisect a finite given Line (A Β.)

Fig. 31.

Upon the given A B make an Equilateral (a) Trian- (a)Per.1.1.1. gle AGB. Bisect its Angle G (b) with the right Line (b) Per praGC. The same shall bisect the given Line A B.

ced.

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

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

PROP. XI. Problem.

FROM

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

a

Perpendicular.

With a Pair of Compasses take the equal Lines AC, AF. From the Centre C and F describe 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 (4) perpendicular to the Line (LI.) (b) Per def.

2. E. F.

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

14.

PROP.