Fig. 33

F

:

PROP. XII. Problem.

ROM a given Point (A) which is without an infinite right Line (as L 2.) to let falt a Perpendicular to that Line.

From the Centre A describe a Circle which may cut the given LQ in C and I. Bisect the right Line CI (c) Per10.1.1.(c) with the right Line A B. This AB is the Perpendicular required.

For let there be drawn AC, AI. Because by the Construction X and Z are equilateral to one another;

(d) Per 8.1.1. Therefore the Angles (d) CBA, IBA, are equal.

(e) Per Def. 14.

Fig. 34.

Therefore, A B is (e) Perpendicular, 2. E. F.

THE

right Line (BA) ftanding upon the right

Line (CF) either makes two right Angles,

Angles equal to two right ones.

or

For if B A stand upon it perpendicularly, then by Definition 14 the two Angles BAC, BAF will be right ones. And if BA stand obliquely, let there be (f) Per11.1,1. rais'd (f) the Perpendicular A L. Where because the unequal Angles CAB, FAB possess the same Place which the two right ones CAL, LAF do, and agree

(g) Per Axi. to them, they are equal (g) to them. 2. E. D.

7.

Fig. 37.

Fig. 36.

Corollaries.

1.IN the fame manner it will be demonftrated, if more right Lines than one stand upon the fame right Line, that the Angles thereby made are equal to two right ones.

2. Two right Lines cutting one another, make the Angles equal to four right ones.

3. All the Angles which are about one Point, make Angles equal to four right ones. It appears from Corollary 2.

4. The Angle CAF being known, you at the fame Fig. 37. time know its Compliment unto two right Angles B A F. For Example, Let the Angle CAF be of 70 Degrees; the Angle BAF will be of 110 Degrees. For those two Numbers added together make 180 Degrees, which is the Measure of two right Angles.

I

PRO P. XIV. Theorem.

F two right Lines (XR, ZR) at the fame Point Fig. 35. of a right Line QR make the Angles on both Sides (XRQ ZRQ) equal to two right Angles; the Lines (XR, ZR) make one right Line.

If you deny it, let XR, BR make one right Line. Therefore the Angles XRQ, QR B (a) will make two (a) Per 13.1. right Angles. Which thing is (b) absurd; seeing by the (b) Contra Hypothefis XRQ, ZRQ do make two right Angles. Axio. 9.

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PROP. XV. Theorem.

F two right Lines (BC, FL) cut one another in A, Fig. 37. the Angles opposite at the top (A) are equal, viz.

LAB to CAF, and BAF to LAC.

For because B A stands upon the right Line LF, the Angles LAB, FAB are (c) equal to two right ones: (c) Per 13.1. And because FA stands upon the right Line BC, the Angles FAC, FAB are also equal (d) to two right (d) By the ones. Therefore the two Angles together (e) LAB, lame Prop. FAB are equal to those two together CAF, FA B. (e)Peraxi.1. Taking away therefore the common Angle FA B, there remains (f) LAB equal to CAF. In the fame manner (f) Peraxi.3. BAF, LACare shewed to be equal.

Coroll. From these two Propositions we gather in Catoptricks, that a Ray of Light, as reflected in an Angle equal to the Angle of Incidence, taketh the shortest way of all.

e. g. When the Angles BEC, AEF are e-Fig. 82. qual, the Lines A E and EB taken together, are shorter than any Lines whatsoever, as AF and FB taken together. For from the Point B let the perpendicular

Line BC be let down; and let BD and DC be equal: Let the Lines also EC and FC be drawn. Now in the Triangles BED and DEC, Seeing the Side DE is common to both, and the Side BD and DC are equal by the Hypothesis, as is also in the like manner BDE equal Per 4.1.1.to the Angle CDE; the Triangles shall be * equal in all other things, and BE shall be equal to CE, and the Angle BED to the Angle DEC: (where because the Angle DEC is equal to [BED, that is] AEF, the Lines A E, EC are proud to make one right Line.) And in the same manner the Line BF will be proved equal to FC. Seeing therefore the Lines B E and FA taken together, are equal to the Line CA, and the Lines BF, FA taken together are equal to the Lines CF, FA taken together; It is manifest that C A, which † Per 20.1.1. is one Side of the Triangle AC Ft, is less than the two Sides CF, FA taken together. Q. E. D.

Fig. 38.

then.

IN

PRO P. XVI, XVII.

HESE two Propositions are contain'd in

PROP. XVIII. Theorem.

Pro

to the greater Side (BO) is the greater; and that (B) which is opposite to the lesser Side (AO) is the leffer Angle.

(A) Cannot be equal to (B) for then the oppofite (2) Per 6.1.1. Sides BO, A O would be equal (a); which is contrary to the Hypothefis. Neither can A be less than B, for if it were so, there might within the Angle B be made an Angle ABF by the right Line BF; which Angle should be equal to A. But then by the 6th of this Book BF, AF shall be equal; and if you add to both OF, then BF, FO shall be equal to AO. But AO by the Hypothefis is less than BO. Therefore BF, FO shall be less than BO, which contradicts the Definition of a right

2

right Line, which is the shortest of all betwixt two Points. Therefore the Angle A is neither less than B, nor equal to it. Therefore it is greater. 2. E. D.

PROP. XIX. Theorem.

N the Triangle AO B the Side (BO) which is op- Fig. 38. the greater Angle (A) is the greater; And that (AO) which is opposed to the lesser Angle B, is the leffer.

:

This Propofition is the Converse of the former. BO is not less than A O, for if it were, the Angle (A) by the 18th would be less than B; which is contrary to the Hypothefis. Nor can BO be equal to A O, for in this Cafe by the 5th, the Angles A and B would be equal. But this Equality of those Angles is contrary to the Hypothefis. Therefore B O is greater than AO. 2. E. D.

Coroll. Hence we gather that a Globe or Ball perfect- Fig. 83. ly polished cannot rest in an horizontal Plane perfectly polished, but where it touches the Earth. For let the Line AB be an horizontal Plane, C the Earth's Centre, CA the Semidiameter of the Earth, perpendicular to the Tangent AB. The Globe placed at B, because of its Gravity, and the Declivity of the Plane, will deScend towards A. For in the Triangle CAB the perpendicular Line C A, which is opposite to the acute Angle ABC, is less than the Line BC which is opposed to the right Angle B AC; and so there is from B to A a perpetual Descent, in which the Globe cannot rest. And in the like manner we prove the Descent of Fluids, and their Conformation into a spherical Surface.

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PROP. XX. Theorem.

N any Triangle, any two Sides of it taken together,
are greater than the remaining Side.

This with Archimedes is as it were an Axiom; forasmuch as it is immediately manifest out of his Definition of a right Line; which see above amongst the Definitions.

Fig. 39.

PROP. XXI. Theorem.

IF from the Ends of one Side AB, two right Lines be drawn, and joined together within the Triangle (as the Lines AO, BO); these are less than the Sides of the Triangle (AC, BC), but they comprehend a greater Angle (AO В).

For as for the first Part of the Propofition, Draw out (a) Per20.1.1. A O unto F: AC, C F are (a) greater than A F. Therefore the common Line F B being added, AC, BC are greater than AF, FB. Again, OF, BF are greater (b) By the (b) than O B. Therefore the common A O being added, A F, BF are greater than AO, BO. Therefore AC, CB are much greater than AO, OB.

fame.

The fecond Part of this Propofition will be demonstrated in the second Corollary of the first Part of Propofition 32. And in the mean while we shall make no ufe of it.

Fig. 40.

T

O make a Triangle of three given right Lines (BO, LB, LO) (of which any two must be

greater than the third.)

Let BL one of the given Lines be taken, and B one of its Extremities being taken for the Centre, with the Interval of the other given Line B O describe an Arch. Then the other Extremity L being taken for the Centre, with the Interval of the third given Line L O defcribe an Arch, cutting the former in O; which being done, and the right Lines BO, LO being drawn, I fay that that is done which was to be done.

The Demonstration is manifest from the Construc