a duplicate proportion of their hamsingen er där mia; wa the squares AĂ, and DM of their bemangan mura

Let the perpendiculars CG and FH be drawn as well as the diagonals BI and EL

The perpendiculars make the triangles ACG and DFH equiangular, and therefore sicer by thec. 16.) for because the angle CAG=FDH and the right angel AGC=DHF, the remaining angle ACG=DFH, (by cor. 2. theo. 5)

Therefore GC: FH::(4C: DF: :) AB: DE, or which is the same thing, GC: AB :: FH: DE for FH multiplied by AB=ABmu.tiplied by FH.

By theo. 19 ABC: ABI :: (CG: 41 or AB as before:: FH DE or DL::) DFE: DLE, therefore ABC: ABI :: DFE : DLE, or ABC: AK:: DFE: DM, for AK is double the triangle ABI, and DM double the triangle DEL, by cor. 2. theo. 12. Q. E. D.

THEO. XXIII

PL. 2. fig. 6.

Like polygons ABCDE, a b c d e, are in a duplicate proportion to that of the sides AB, a b, which are between the equal angles A and B and a and b, or as the aquarts of the sides AB, ab.

Draw AD, AC, ad, ac.

By the hypothesis AB: ab:: BC: bc, and thereby also the angle B=6; therefore (by theo. 21.) BAC-bac; and ACB=acb: in like manner EAD-e ad, and ED4➡e da. If therefore from the equal angles A, and a, we take the equal ones

EAD+BAC e ad,+bac the remaining angle DAC=d a c, and if from the equal angles D and d, EDA=ed a, be taken, we shall have ADC=ad c and in like manner if from C and c be taken BCA-bca, we shall have ACD-a cd; and so the respective angles in every triangle, will be equal to those in the other.

By theo. 22. ABC: abc the square of AC to the square of ac, and also ADC: adc: the square of AC, to the square of a c; therefore from equality of proportions ABC: abc:: ADC: a dc; in like manner we may shew that ADC: ad c: EAD: ead: Therefore it will be as one antecedent is to one consequent, so are all the antecedents to all the consequents. That is, ABC is to a b c as the sum of the three triangles in the first polygon, is to the sum of those in the last. Or ABC will be to a b c, as polygon to polygon.

The proportion of ABC to abc (by the foregoing theo.) is as the square of AB is to the square of a b, but the proportion of polygon to polygon, is as ABC to a b c, as now shown: therefore the proportion of polygon to polygon is as the square of B to the square of ab.

THEO. XXIV.

PL. 1. fig. 8.

Let DHB be a quadrant of a circle described by the radius CB; HB an arc of it, and DH its complement ; HL or FC the sine, FH or CL its co-sinę, BK its tangent, DI its cotangent; CK its secant, and CI its co-secant. Fig. 8.

1. The co-sine of an arc is to the sine, as the radius is to the tangent.

2. The radius is to the tangent of an arc, as the co-sine of it is to the sine.

3. The sine of an arc is to its co-sine, as the radius to its co-tangent;

4. Or the radius is to the co-tangent of an arc, as its sine to its co-sine.

5. The co-tangent of an arc is to the radius, as the radius to the tangent.

6. The co-sine of an arc is to the radius, as the radius is to the secant.

7. The sine of an arc is to the radius, as the tangent is to the secant.

The triangles CLH and CBK, being similar, (by theo. 16.)

1. CL: LH:: CB: BK.

2. Or, CB: BK: : CL: LH.

The triangles CFH and CDI, being similar.

3. CF (or LH): FH:: CD: DI.

4. GD: DI:: CF (or LH): FH.

The triangles CDI and CBK are similar for the angle CIP-KCB, being alternate ones (by part 2. theo. 3.) the lines CB and D being parallel the angle CDI=CBK being both right, and consequently the angle DCI-CKB, wherefore,

5. DI: CD:: CB: EK.

And again, making use of the similar triangle CLHand CBK.

6. CL: CB:: CH: CK.

7. HL:CH: BK : CK,

GEOMETRICAL PROBLEMS.

PROB. I.

PL. 2. fig. 7.

To make a triangle of three given right lines BO, LB, LO, of which any two must be greater than the third.

Lay BL from B to L; from B with the line BO, describe an arc, and from L with LO describe another arc; from O, the interescting point of those arcs, draw BO and OL, and BOL is the triangle required.

This is manifest from the construction.

PROB. II.

PL. 2. fig. 8.

At a point B in a given right line BC, to make an angle equal to a given angle A.

Draw any right line ED to form a triangle, as EAD, take BF-AD, and upon BF make the triangle BFG, whose side BG=AE, and GF=ED (by the last) then also the angle B =A; if we suppose one triangle be laid on the other, the sides

will mutually agree with each other, and therefore be equal; for if we consider these two triangles to be made of the same three given lines, they are manifestly one and the same triangle.

Otherwise.

Upon the centres A and B, at any distance, let two arcs, DE, FG, be described; make the are FG=DE, and through Band G draw the line BG, and it is done.

For since the chords ED, GF, are equal, the angles A and B are also equal, as before (by def. 17.)

PROB. III.

PL. 2. fig. 9.

To bisect or divide into two equal parts, any given rightlined angle, BAC.

=

In the lines AB and AC, from the point A set off equal distances AE, AD, then, with any distance more than the half of DE, describe two arcs to cut each other in some point F; and the rightline AF, joining the points A and F, will bisect the given angle BẮC.

For if DF and FE be drawn, the triangles ADF, AEF, are equilateral to each other viz. AD =AE, DF-FE, and AF common, wherefore DAF-EAF, as before.

PROB. IV.

PL. 2. fig. 10.

To bisect a right-line. AB.

With any distance, more than half the line, from

K