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BOOK IV.

PROP. I. PROB.

Given the base of a triangle, the difference of the sides, and the difference of the angles at the base ; to construct it.

A

D

Make BC equal to the given base, and CBD equal to half the difference of the angles at the base : from C as centre, at a distance equal to the difference of the sides, describe an arc cutting BD in D: join CD and produce it: make the angle DBA equal to BDA: ABC is the required triangle.

For (I. 6.) AD is equal to AB, and the difference of AC, AD, or of AC, AB, is CD; and (App. I. 6.) since AD is equal to AB, CBD is equal to half the difference of the angles at the base. The triangle ABC, therefore, is the one required, as it has its base equal to the given base, the difference of its sides equal to the given difference, and the difference of the angles at the base equal to the given difference of those angles.

Method of Computation. In the triangle BCD, there are given BC, CD, and the angle CBD; whence the angle C can be computed ; and the sum of this and twice CBD is the angle ABC. Then, in the whole triangle ABC, the angles and the side BC are given ; whence the other sides may be computed : or, one of them being computed, the other will be found by means of the given difference CD.

B

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PROP. II. PROB.

Given the segments into which the base of a triangle is divided by the line bisecting the vertical angle, and the difference of the sides; to construct the triangle.

Construct the triangle CED, having the sides CE, ED equal to the given segments, and CD equal to the given difference of the sides : produce CE, and make EB equal to ED ; bisect the angle BED by EA, meeting CD produced in A, and join AB : ABC is the required triangle.

For, in the triangles AEB, AED, BE is equal to ED, EA common, and the angle BEA equal to DEA: therefore (I. 4.) BA is equal to DA, and the angle EAB to EAD. Hence ABC is the required triangle; for CD, the difference of its sides, is equal to the given difference, and BE, EC, the segments into which the base is divided by the line bisecting the vertical angle, are equal to the given segments.

Method of Computation. The sides of the triangle CDE are given, and therefore its angles may be computed: one of which, and the supplement of the other are the angles C and B. Then, in the triangle ABC, the angles and BC are given, to compute the remaining sides.

A

B

E

OTHERWISE. Since (VI. 3.) CE: EB :: CA : AB, we have, by division, CE-EB: EB :: CA- AB: AB; which, therefore, becomes known, since the first three terms of the analogy are given ; and thence AC will be found by adding to AB the given difference of the sides.

PROP. III. PROB.

M

Given the base of a triangle, the vertical angle, and the difference of the sides ; to construct the triangle.

Let MNO be the given vertical angle; produce ON to P, and bisect the angle MNP by NQ. Then, make BD equal to the difference of the sides, and the angle ADC equal to QNP; from B as centre, with the given base as radius, describe an arc cutting DC in C; and make the angle DCA equal to ADC: ABC is the required triangle. For (const.) the angles ACD, ADC are equal to MNP, and therefore

P

D

B

(1. 32. and 13.) the angle A is equal to MNO. But (I. 6.) AD is equal to AC, and therefore BD is the difference of the sides AB, AC; and the base BC is equal to the given base: wherefore ABC is the triangle required.

Method of Computation. In the triangle CBD, the sides BC, BD, and the angle BDC, the supplement of ADC or MNQ are given ; whence the other angles can be computed. The rest of the operation will proceed as in the first proposition of this book.

D

PROP. IV. PROB. Given one of the angles at the base of a triangle, the difference of the sides, and the difference of the segments into which the base is divided by the line bisecting the vertical angle; to construct the triangle.

Construct the triangle DBG, having the angle B equal to the given angle, BD equal to the difference of the sides, and BG equal to the difference of the segments ; draw DC perpendicular to DG; produce BD; and make the angle DCA equal to CDA: ABC is the triangle required. For it has B equal to the given angle, and the difference of its sides BD equal to the given difference: and if AHE be drawn bisecting the angle BAC, it bisects (I. 4.) CD, and is perpendicular to it: it is therefore parallel to DG, one side of the triangle CDG; and bisecting CD in H, it also (VI. 2.) bisects GC in E. Hence BG, the difference of BE, GE, is also the difference of BE, EC, the segments into which the base is divided by the line bisecting the vertical angle.

Method of Computation. In the triangle DBG, BD, BG, and the angle B are given ; whence (APP. III. 3.) we find half the difference of the angles BGD, BDG, which is equal to balf the angle C. *

Then (by the same proposition) we have, in the triangle ABC, tan (C— B) : tant (C +B) ::6-b or BD : c+b; whence the sides c and 6 become known, and thence BC by the first case.

B

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PROP. V. PROB. Given the base of a triangle, the vertical angle, and the sum of the sides ; to construct it.

Make BD equal to the sum of the sides, and the angle D equal

* For (I. 32.) BEA=A+Cand consequently BEA-A= C. But (I. 29.) BGD= BEA, and BDG= BAE=A; and therefore BGD-BDG = C.

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A

B

to half the vertical angle; from B as centre, with the base as radius, describe an arc meeting DC in C;* and make the angle DCA equal to D; ABC is the required triangle.

For (1. 6.) AD is equal to AC, and therefore BA, AC are equal to BD, the given sum. Also (I. 32.) the exterior angle BAC is equal to the two D and ACD, or to the double of D, because D and ACD are equal : therefore, since D is half the given vertical angle, BAC is equal to that angle. The triangle ABC, therefore, bas its base equal to the given base, its vertical angle equal to the given one, and the sum of its sides equal to the given sum: it is therefore the triangle required.

Method of Computation. In the triangle BDC, the angle D, and the sides BC, BD are given : whence the remaining angles can be computed: and then, in the triangle ABC, the angles and the side BC are given, to compute the other sides.

PROP. VI. PROB.

Given the vertical angle of a triangle, and the segments into which the line bisecting it divides the base ; to construct it.

A

In the straight line BC, take BH and CH equal to the segments of the base ; on BC describe (III. 33.) the segment BAC containing an angle equal to the vertical angle, and complete the circle ; bisect the arc BEC in E; draw EHA, and join BA, CA: ABC is the required triangle. For (III. 27.) the angles BAH, CAH are equal, because the arcs BE, EC are equal: and therefore the triangle ABC manifestly answers the conditions of the question. +

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D

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* Should the circle neither cut nor touch DC, the problem would be impossible with the proposed data. If the circle meet DC in two points, there will be two triangles each of which will answer the conditions of the problem. These triangles, however, will differ only in position, as they will be on the same base, and will have their remaining sides equal, each to each. This problem might also be solved by describing (III. 33.) on the given base BC a segment of a circle containing an angle equal to half the vertical angle; by inscribing a chord BD equal to the sum of the sides ; by joining DC; and then proceeding as before. The construction given above is preferable.

† The construction might also be effected by describing on BH and CH segments each containing an angle equal to half the vertical angle, and joining their point of intersection A with B and C. Another solution may be obtained

Method of Computation. Join BE, and draw ED perpendicular to BC. Then BD or DC is half the sum of the segments BH, HC, and DH half their difference: and BD is to DH, or twice BD to twice DH, as tan DEB to tan DEH. Now it is easy to show that BED is half the sum of the angles ABC, ACB, and HED half their difference; and therefore these angles become known; and BC being given, the triangle ABC is then resolved by the method for the first case.

Cor. Hence we have the method of solving the problem in which the base, the vertical angle, and the ratio of the sides of a triangle are given, to construct it. For (VI. 3.) the sides being proportional to the segments BH, HC, it is only necessary to divide the given base into segments proportional to the sides, and then to proceed as above.

PROP. VII. PROB.

Given the base, the perpendicular, and the vertical angle of a triangle; to construct it.

Make BC equal to the given base, and (III. 33.) on it describe a segment capable of containing an angle equal to the vertical angle; draw AG parallel to BC, at a distance from it equal to the given perpendicular, and meeting the arc in A; join AB, AC: ABC is evidently the triangle required.

Method of Computation. Draw the perpendicular AD, and parallel to it draw LGHK through the centre F; join BF, AF, AK. Now, since AK evidently bisects the angle BAC, the angle KAD or K is (APP. I. 6.) equal to half the difference of the angles ABC, ACB, and therefore (III. 20.) AFG is the whole difference of those angles. Then, in the right-angled triangle BHF, the angles and BH being known, FH can be computed ; from which and from AD or GH, FG becomes known. Now, to the radius BF or AF, FH is the cosine of BFH or BAC, and FG the cosine of AFG; and therefore FH :

B

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D

by the principle (VI. 3.) that BA : AC :: BH : HC. For if a triangle be constructed baving its vertical angle equal to the given one, and the sides containing it equal to the given segments, or having the same ratio, that triangle will be similar to the required one ; and therefore on AB construct a triangle equiangular to the one so obtained.

A fourth solution may be obtained from proposition G. of the sixth book; but it is not so easy as those already pointed out.

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