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PROP. IV.

THE fum of any two fides of a triangle is to their difference, as the tangent of half the fum of the angles oppofite to those fides, to the tangent of half their difference.

Let ABC be any plain triangle;

CA+AB: CA-AB: ; tan.

(B+C): tan. (B–C).

For (2.) CA: AB :: fin. B; fin. C;

and therefore (E. 5.)

CA + AB: CA-AB :: fin. B+fin. C: fin. B — fin. C.
But, by the laft, fin. B+fin. C: fin. B — fin. C ::

tan. (B+C): tan (B-C); therefore also, (11. 5.) CA+AB: CA-AB:: tan.

(B+C): tan ÷ (B—C). Q

E. D.

A

B

C

PROP.

303

Otherwise, without the 3d.

Let ABC be a triangle; the fum of AB and AC fides, is to the difference of AB and AC, as the tangent of any two half the fum of the angles ACB and ABC, to the tangent of, half their difference.

About the centre A with the radius AB, the greater of

the two fides, describe a circle meeting BC produced in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G.

Because the exterior angle EAB is equal to the two interior ABC, ACB, (32. 1.); and the angle EFB, at the circumference is e

qual to half the angle

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EAB at the centre (20. 3.); therefore EFB is half the fum of the angles oppofite to the fides AB and AC.

Again, the exterior angle ACB is equal to the two interior ČAD, ADC, and therefore CAD is the difference of the angles ACB, ADC, that is of ACB, ABC, for ABC is equal to ADC. Wherefore alfo DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles oppofite to the fides AB, A C.

Now because the angle FBE in a femicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFG to the radius FB; and BE is therefore to BG as the tangent of half the fum of the angles ACB, ABC to the tangent of half their difference. Alfo CE is the fum of the fides of the triangle ABC, and CF

their

their difference; and because BC is parallel to FG, CE: CF:: BE: BG, (2. 6.); that is, the fum of the two fides of the triangle ABC is to their difference as the tangent of half the fum of the angles oppofite to thofe fides to the tangent of half their difference. Q. E. D.

PROP. V.

F a perpendicular be drawn from any angle of a

I fide,

the fegments of the bafe is to the fum of the other two fides of the triangle as the difference of those fides to the difference of the fegments of the bafe.

For (K. 6.), the rectangle under the fum and difference of the fegments of the bafe is equal to the rectangle under the fum and difference of the fides, and therefore (16.6.) the sum of the fegments of the base is to the fum of the fides as the difference of the fides to the difference of the fegments of the bafe. Q. E. D.

IN

PROP. VI.

N any triangle, twice the rectangle contained by any two fides is to the difference between the fum of the fquares of thofe fides, and the fquare of the bafe, as the radius to the cofine of the angle included by the two fides.

Let

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COR. If the radius 1, BD BAX cof B, (1.), and 2BC.BA x cof B2BC.BD, and therefore when B is acute, 2BC.BA X cof BBC2+ BA2-AC2, and adding AC2 to both; AC2+2 cof B x BC.BA BC2+ BA2; and taking 2 cof BXBC.BA from both, AC2=BC2-2 cof BXBC.BA+ BA2. Wherefore AC (BC2-2 cof B x BC.BA+BA2). If B is an obtufe angle, it is fhewn in the fame way that AC=√(BC2 + 2 cof B × BC.BA + BA2.).

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PROP. VII.

OUR times the rectangle contained by any twe fides of a triangle, is to the rectangle contained by two ftraight lines, of which one is the bafe or third fide of the triangle increafed by the dif ference of the two fides, and the other the bafe diminished by the difference of the fame, fides, as the fquare of the radius to the fquare of the fine of half the angle included between the two fides of the triangle.

Let ABC be a triangle, of which BC is the base, and AB the greater of the two fides: 4AB.AC: (BC+(AB-AC))x (BC — (AB -- AC)) : R2 : (fin 1⁄2 BAC)2.

Produce the fide AC to D, fo that ADAB; join BD, and draw AE,CF at

right angles to it; from the centre C with the radius CD defcribe the femicircle DGH, cutting BD in K, BC in G, and meeting BC produced in H. It is plain that CD is the difference of the fides,

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and therefore that BH is the bafe increased, and BC the base diminished by the difference of the fides; it is alfo evident, because the triangle BAD is ifofceles, that DE is the half of BD, and DF is the half of DK, wherefore DE-DF the half of BD-DK, (6. 5.) that is EF BK. And because AE is drawn parallel to CF, a fide of the triangle CFD, AC: AD: : EF: ED, (2. 6.); and rectangles of the fame altitude being as their bases AC.AD: AD2:: EF.ED: ED, ( 6.), and therefore 4AC.AD: AD2:: 4EF.ED: ED, or alternately, 4AC.AD: 4EK.ED : : AD2 : ED2.

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