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* 50 Dat.

con* to AF, so is* the rectangle DC, CE, to the rectangle * 1. 6. AF, CE; and the ratio of the rectangle AF, CE, to its half *, the triangle ACE, is given; therefore the ratio * 41. 1. of the rectangle DC, CE, to the triangle ACE, that is*, * 37. 1. to the triangle ABC is given*; and the rectangle DC, * 9 Dat. CE, is the excess of the square of BA, AC, together, above the square of BC, therefore the ratio of this excess to the triangle ABC is given.

The ratio which the rectangle DC, CE, has to the triangle ABC is found thus: take the straight line GH given in position and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join KH, and draw GL perpendicular to it: then the ratio of HK to the half of GL is the same with the ratio of the r

recta angle DC, CE, to the triangle ABC: because the angles HGK, DAC, at the vertices of the isosceles triangles GHK, ADC, are equal to one another, these triangles are similar; and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so is* *4. 6. (DC to AF, and so is) the rectangle DC, CE, to the rectangle AF, CE; but as GE to its half

, so is the rectangle AF, CE, to its half, which is the triangle ACE, or the

triangle ABC; therefore ex æquali, HK is to the half of the straight line GL, as the rectangle DC, CE, is to the triangle ABC.

Cor. And if a triangle have a given angle, the space by which the square of the straight line, which is the difference of the sides which contain the given angle, is less than the square of the third side, shall have a given ratio to the triangle. This is demonstrated the same way as in the preceding proposition, by help of the second case of the Lemma.

22. 5.

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If the perpendicular drawn from a given angle of a See N.

triangle to the opposite side, or base, have a given ratio to the base, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC, have a given ratio to it, the triangle ABC is given in species. egerit.

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If ABC be an isosceles triangle, it is evident*, that if any one of its angles be given, the rest are also

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given; and therefore the triangle is given in species, without the consideration of the ratio of the perpendicular to the base which in this case is given by Prop. 50.

But when ABC is not an isosceles triangle, take any straight line EF given in position and magnitude, and upon it describe the segment of a circle EGF, containing an angle equal to the given angle BAC, draw GH bisecting EF at right angles, and join EG, GF: then, since the angle EGF is equal to the angle BAC, and that EGF is an isosceles triangle, and ABC is not, the angle FEG is not equal to the angle CBA: draw EL, making the angle FEL, equal to the angle CBA; join FL, and draw LM perpendicular to EF; then because the triangles ELF, BAC, are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB: and as LE to EF, so is AB to BC; wherefore, ex æquali, as LM to EF, so is AD to BC; and because the ratio of AD to BC is given, therefore the ratio of LM to EF is given ; and EF is given, wherefore* LM also is given. Complete the parallelogram LMFK; and because LM is given, FK is given in magnitude; it is also given in position; and the point F is given, and consequently* the point K; and because through K the straight line KL is drawn parallel to EF, which is given in position, therefore* KL is given in position: and the circumference ELF is given in position; therefore the point L is given *: and because the points L, E, F, are given,

the straight lines LE, EF, FL, are given* in magni* 42 Dat. tude; therefore the triangle LEF is given in species *;

and the triangle ABC is similar to LEF, wherefore also ABC is given in species.

Because LM is less than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be less

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than the ratio of GH to EF, which the straight line in a segment of a circle containing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base.

Cor. 1. If two triangles, ABC, LEF, have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the base BC, as the perpendicular LM to the base EF, the triangles, ABC, LEF, are similar.

Describe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicular to EF; because the angles ENF, ELF, are equal, and that the angle EFN is equal to the alternate angle FNL, that is, to the angle FEL, in the same segment; therefore the triangle NEF is similar to LEF; and in the segment EGF there can be no other triangle upon the base EF, which has the ratio of

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its perpendicular to that base the same with the ratio of LM or NO to EF, because the perpendicular must be greater or less than LM or NO; but, as has been shewn in the preceding demonstration, a triangle, similar to ABC, can be described in the segment EĞF, upon the base EF, and the ratio of its perpendicular to the base is the same, as was there shewn, with the ratio of AD to BC, that is, of LM to EF; therefore that triangle must be either LEF, or NEF, which therefore are similar to the triangle ABC.

Cor. 2. If a triangle ABC have a given angle BAC, and if the straight line AR drawn from the given angle to the opposite side BC, in a given angle ARC, have a given ratio to BC, the triangle ABC is given in species.

Draw AD perpendicular to BC; therefore the triangle ARD is given in species; wherefore the ratio of AD to AR is given: and the ratio of AR to BC is given, and consequently* the ratio of AD to BC is * 9 Dat.

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22. 5.

given; and the triangle ABC is therefore given in * 77 Dat. species*.

CoR. 3. If two triangles ABC, LEF,_have one angle BAC equal to one angle ELF, and if straight lines drawn from these angles to the bases, making with them given and equal anglés, have the same ratio to the bases, each to each; then the triangles are similar: for having drawn perpendiculars to the bases from the equal angles, as one perpendicular is to its base, so is the other to its base*; wherefore, by Cor. 1. the triangles are similar.

A triangle similar to ABC may be found thus : having described the segment EGF, and drawn the straight line GH as was directed in the proposition, find FK, which has to EF the given ratio of AD to BC; and place FK at right angles to EF from the point F; then because, as has been shewn, the ratio of AD to BC, that is, of FK to EF, must be less than the ratio of GH to EF, therefore FK is less than GH; and consequently the parallel to EF drawn through the point K, must meet the circumference of the segment in two points: let L be either of them, and join EL, LF, and draw LM perpendicular to EF: then, because the angle BAC is equal to the angle ELF, and that AD is to BC, as KF, ihat is, LM, to EF, the triangle ABC is similar to the triangle LEF, by Cor. 1.

80.

PROP. LXXVIII.

If a triangle have one angle given, and if the ratio

angle to the square of the third side be given,
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the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AČ, to the square of BC be given; the triangle ABC is given in species.

From the point A, draw AD perpendicular to BC, the rectangle AD, BC, has a given ratio to its half* the triangle ABC; and because the angle BAC is

given, the ratio of the triangle ABC to the rectangle * Cor. 62. BA, AC is given*, and by the hypothesis, the ratio

of the rectangle BA, AC, to the square of BC is given;

* 41. 1.

Dat.

* 77 Dat.

therefore * the ratio of the rectangle AD, BC, to the *9 Dat. square of BC that is *, the ratio of the straight line *1. 6. AD to BC, is given ; wherefore the triangle ABC is given in species *

A triangle similar to ABC may be found thus: take a straight line EF given in position and magnitude, and make the angle FEG equal to the given angle BAC, and draw FH perpendicular to EG, and BK perpendicular to AC; therefore the

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A. triangles ABK,

K

L EFH, are similar, and the rectangle

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H AD, BC, or the

R rectangleBK, AC

B D N which is equal

F to it, is to the rectangle BA, AC, as the straight line BK to BA, that is, as FH to FE. Let the given ratio of the rectangle BA, AC, to the square of BC, be the same with the ratio of the straight line EF to FL; therefore, ex æquali, the ratio of the rectangle AD, BC, to the square of BC, that is, the ratio of the straight line AD to BC, is the same with the ratio of HF to FL; and because AD is not greater than the straight line MN in the segment of the circle described about the triangle ABC, which bisects BC at right angles ; the ratio of AD to BC, that is, of HF to FL, must not

greater than the ratio of MN to BC: let it be so; and by the 77th Dat. find a triangle OPQ which has one of its angles POQ equal to the given angle BAC, and the ratio of the perpendicular OR, drawn from that angle to the base PQ, the same with the ratio of HF to FL; then the triangle ABC is similar to OPQ: because, as has been shewn, the ratio of AD, to BC is the same with the ratio of (HF to FL, that is, by the construction, with the ratio of) OR to PQ; and the angle BAC is equal to the angle POQ. Therefore the triangle ABC is similar * to the triangle POQ.

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1 Cor. 77 Dat.

Otherwise. Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC, to the square of BC, be given; the triangle ABC is given in species.

Because the angle BAC is given, the excess of the square of both the sides BA, AC, together, above the

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