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a given ratio to one another; the triangle ABC is given in species.

Describe the circle BAC about the triangle, and from its centre E, draw EA, EB, EC, ED; becaufe the angle BAC is given, the angle BEC at the centre, which is the double b of it, is given. And the ratio of BE to EC is given, because they are equal to one another ; therefore the triangle BEC is

given in species, and the ratio of EB to BC given ; also the d 7. dat.

ratio of CB to BD is given , because the ratio of BD to DC cg. dat. is given ; therefore the ratio of EB to BD is given, and the

angle EBC is given, wherefore the triangle EBD is given in fpecies, and the ratio of EB, that is, of EA to ED, is therefore given; and the angle EDA is given, because each of the

angles BDE, BDA is given ; therefore the triangle AED is f 49. dat. given in species, and the angle AED gi

ven; also the angle DEC is given, be-
cause each of the angles BED, BEC is
given ; therefore the angle AEC is given,
and the ratio of EA to EC, which are
equal, is given ; and the triangle AEC is

B

C therefore given in fpecies, and the angle

D
ECA given; and the angle ECB is given,

wherefore the angle ACB is given, and the angle BAC is also 8 43. dat. given ; therefore the triangle ABC is given in species.

A triangle similar to ABC may be found, by taking a straight line given in position and magnitude, and dividing it in the given ratio which the segments BD, DC are required to have to one another; then, if upon that straight line a fegment of a circle be described containing an angle equal to the given angle BAC, and a straight line be drawn from the point of division in an angle equal to the given angle ADB, and from the point where it meets the circumference, straight lines be drawn to the extremity of the first line, there, together with the firft line, shall contain a triangle similar to ABC, as may easily be shown.

The demonstration may be also made in the manner of that of the 77th prop. and that of the 77th may be made in the manner of this.

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F the fides about an angle of a triangle have a głven

ratio to one another, and if the perpendicular drawn from that angle to the base has a given ratio to the base ; the triangle is given in species.

Let 43

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43. dat.

Let the Gides BA, AC, about the angle BAC of the triang'e ABC have a given ratio to one another, and let the perpendicua lar AD have a given ratio to the base BC; the triangle ABC is. given in species.

First, let the Gides AB, AC be equal to one another, therefore the perpendicular AD bifects a the base A

a 26. 1. BC; and the ratio of AD to BC, and there. fore to its half DB, is given ; and the angle 1 ADB is given ; wherefore the triangle * ABD, and consequently the triangle ABC, is given 6 B D C 644. dar. in species.

But let the Gides be unequal, and BA be greater than AC; and make the angle CAE equal to the angle ABC ; because the angle AEB is common to the triangles ACB, CEA, they are fimilar; therefore as AB to BE, so is CA to AE, and, by permutation, as BA to AC, so is BE to EA, and so is EÁ to EC ; and the ratio of BA to AC is given, therefore the ratio of BE to EA, and the ratio of EA to EC, as also the ratio of BE to EC is given; wherefore the ratio of EB to c 9. dat. BC is given d; and the ratio of AD to BC

A d 6. dat. is given by the hypothesis, therefore the ratio of AD to BẾ is given; and the ratio 2. of BE to EA was shown to be given ; wherefore the ratio of AD to AE is given, and B FC E D ADE is a right angle, therefore the triangle ADE is given in species, and the angle AEB given; the ra-c 46. dat. tio of BE to EA is likewise given, therefore the triangle ABE is given in species, and consequently the apgle EAB, as also the angle ABE, that is, the angle CAE, is given ; therefore the angle BAC is given, and the angle ABC being also given, the triangle ABC is given f in species.

f 43. dat. How to find a triangle which shall have the things which are mentioned to be given in the propofition, is evident in the first case; and to find it the more easily in the other cafe, it is to be observed that, if the straight line EF equal to EA be placed in EB towards B, the point F divides the base BC into the segments BF, FC which have to one another the ratio of the fides BA, AC; because BE, EA, or EF, and EC were shown to be proportionals, therefore * BF is to FC, 19. s. as BE to EF, or EA, that is, as BA to AC; and A£ cannot be less than the altitude of the triangle ABC, but it may be Еe 2

equal

6.2.

equal to it; which, if it be, the triangle, in this case, as also
the ratio of the fides, may be thus found ; having given the
ratio of the perpendicular to the base. Take the straight line
GH given in position and magnitude, for the base of the tri-
angle to be found; and let the given ratio of the perpendicu-
lar to the base be that of the straight line K to GH, that is,
let ķ be equal to the perpendicular; and suppose GLH to be
the triangle which is to be found, therefore having made the
angle HLM equal to LGH, it is required that LM be per-
pendicular to GM, and equal to K; and because GM, ML,
MH are proportionals, as was shown of BE, EA, EC, the
rectangle GMH is equal to the square of ML. Add the com.
mon square of NH, (having bisected GH in N), and the square
of NM is equal to the squares of the given straight lines NH
and ML, or K; therefore the square of NM, and its fide
NM, is given, as also the point M, viz. by taking the straight
line NM, the square of which is equal to the squares of NH,
ML. Draw ML equal to K, at right angles to GM; and be-
cause ML is given in position and magnitude, therefore the
point L is given ; join LG, LH; then the triangle LGH is
that which was to be found, for the square of NM is equal to
the squares of NH and ML, and taking away the common
square of NH, the rect-
angie GMH is equal to K L R
the square of ML; there-

S
fore as GM to ML, so is 3.
ML to MH, and the tri-
angle LGM is therefore
equiangular to HLM, and
the angle HLM equal to G NQH

NQ H MP the angle LGM, and the straighi line LM, drawn from the vertex of the triangle making the angle HLM equal to LGH, is perpendicular to the base and equal to the given straight line K, as was required ; and the ratio of the sides GL, LH is the same with the ratio of GM to ML, that is, with the ratio of the straight line which is made

of GN the half of the given base and of NM, the square of which is equal to the squares of GN and K, to the straight line K.

And whether this ratio of GM to ML is greater or less than the ratio of the sides of any other triangle upon the base GH, and of which the altitude is equal to the Atraight line K,

the

li 6. 6.

up

that is, the vertex of which is in the parallel to GH drawn through the point L, may be thus found. Let OGH be any such triangle, and draw OP, making the angle HOP equal to the angle OGH; therefore, as before, GP, PO, PH are proportionals, and PO cannot be equal to LM, because the rect. angle GPH would be equal to the rectangle GMH, which is impossible ; for the point P cannot fall upon M, because () would then fall on L; nor can PO be less than LM, therefore it is greater; and consequently the rectangle GPH is greater than the rectangle GMH, and the Itraight line GP greater than GM: Therefore the ratio of GM to MH is greater than the ratio of GP to PH, and the ratio of the square of GM to the square of ML is therefore i greater than the ratio of the i 2. Cor. square of GP to the square of PO, and the ratio of the straight 20. 6. line GM to ML greater than the ratio of GP to PO. But as GM to ML, so is GL to LH; and as GP to PO, so is GO to OH; therefore the ratio of GL to LH is greater than the ratio of GO to OH; wherefore the ratio of GL to LH is the greateft of all others; and consequently the given ratio of the greater lide to the less must not be greater than this ratio.

But if the ratio of the fides be not the same with this great est ratio of GM to ML, it must necessarily be less than it: Let any less ratio be given, and the same things being suppos fed, viz. that GH is the base, and K equal to the altitude of the triangle, it may be found as follows. Divide GH in the point Q, so that the ratio of GQ to QH may be the same with the

given ratio of the sides; and as GQ to QH, so make GP to PQ, and so will f PO be to PH; wherefore the square ? 19. s. of GP is to the square of PQ, as i the straight line GP to PH: And because GM, ML, MH are proportionals, the square of GM is to the square of ML, as i the straight line GM to MH: But the ratio of GQ to QH, that is, the ratio of GP to PQ, is less than the ratio of GM 10 ML; and therefore the ratio of the square of GP to the square of PQ is less than the ratio of the Iquare of GM to that of ML; and consequently the ratio of the straight line GP to PH is less than the ratio of GM to MH; and, by division, the ratio of GH to HP is less than that of GH to HM; wherefore " the ftraight line HP is k 10. so greater than HM, and the rectangle GPH, that is, the square of PQ, greater than the rectangle GMH, that is, than the

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square of ML, and the straight line PQ is therefore greater than ML. Draw LR parallel to GP, and from P draw PR at sight angles to GP: Because PQ is greater than ML, or PR, the circle described from the centre P, at the distance PQ, must necessarily cut LR in two points ; let these be O, S, and join OG, OH; SG, SH; each of the triangles OGH, SGH have the things mentioned to be given in the proposition : Join OP, SP, and because as GP to PQ, or PO, lo is PO to PH, the triangle OGP is equiangular to HOP; as, therefore, OG to GP, fo is HO to OP, and, by permutation, as GO to OH, so is GP to PO, or PQ; and so is GQ to QH: Therefore the triangle OGH has the ratio of its fides GO, OH the same with the given ratio of GQ to QH; and the perpendicular has to the base the given ratio of K to GH, because the perpendicular is equal to LM, or K: The like may be shewn in the same way of the triangle SGH.

This construction by which the triangle OGH is found, is shorter than that which would be deduced from the demon. itration of the datum, by reason that the base GH is given in position and magnitude, which was not suppofed in the demonstration : The same thing is to be observed in the next proposition.

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[F the sides about an angle of a triangle be unequal

and have a given ratio to one another, and if the perpendicular from that angle to the base divides it into segments that have a given ratio to one another, the triangle is given in fpecies.

Let ABC be a triangle, the fides of which about the angle BAC are unequal, and have a given ratio to one another, and let the perpendicular AD to the base BC divide it into the seg. ments BD, DC which have a given ratio to one another, the triangle ABC is given in fpecies.

Let AB be greater than AC, and make the angle CAE equal to the angle ABC, and because the angle AEB is common to the triangles ABE, CAE, they are equiangular to one another : Therefore as AB to BE, so is CA to AE, and,

by

24. 6,

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