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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 BÅH, CAH are equal, because the arcs BE, EC are equal : and therefore the triangle ABC manifestly answers the conditions of the question.*

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 ÅG 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

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* 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 by the principle (VI. 3) that BA : AC :: BH : HC. For if a triangle be constructed having 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 CB 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.

ACB, and therefore (III. 20) AFG is the whole difference of those angles. Then, in the right-angled triangle BHF, the angles and BĦ 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 :FG :: cosBAC : COSAFG. Hence the angles ABC, ACB become known, and thence the remaining sides.

Schol. Should the parallel AG not meet the circle, the solution would be impossible, as no triangle could be constructed having its base, perpendicular, and vertical angle of the given magnitudes. If the parallel cut the circle, there will be two triangles, either of which will answer the conditions of the question. They will differ, however, only in position, as their sides will be equal, each to each. If the parallel touch the circle, there will be only one triangle; and it will be isosceles.

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Prop. VIII. PROB.--Given the base of a triangle, the vertical angle, and the radius of the inscribed circle; to construct the triangle.

Let GHK be the given angle; produce GH to L, and bisect LHK by HM: on the given base BC describe (III. 33) the segment BDC, containing an angle equal to GHM: draw a straight line parallel to BC, at a distance equal to the given radius, and meeting the arc of the segment in D: join DB, DC; and make the angles DBA, DCA equal to DBC, DCB, each to each: ABC is the required triangle.

Produce BD to E, and draw DF perpendicular to BC. Then, since (const.) the angle BDC is equal to GHM, the two DBC, DCB are equal (I. 32 and 13) to LHM, and therefore (const.) ABD, ACD are equal to KHM. But (I. 32) BDC is equal to BEC, ECD, orto BAC, ABD, ATD, because (I. 32) BEC is equal to BAC, ABD. Therefore BAC, ABD, ACD are equal to GHM: from the former take ABD, ACD, and from the latter KHM, which is equal to them, and the remainders BAC, GHK are equal. It is plain also (I. 26) that perpendiculars drawn from D to AB and 'AC, would be each equal to DF ; and therefore a circle described from D as centre, with DF as radius, would be inscribed in the triangle ABC; and BC being the given base, and A being equal to the given vertical angle, ABC is the required triangle.

The method of computation is easily derived from that of the preceding proposition.

Schol. Should the parallel to BC not meet the arc of the segment, the solution would be impossible, as there would be no triangle which could have its base, its vertical angle, and its inscribed circle of the given magnitudes. If the parallel be a tangent to the arc, the radius of the inscribed circle would be a maxi

Hence, to solve the problem in which the base and the

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vertical angle are given, to construct the triangle, so that the inscribed circle may be a maximum, describe the segment as before, and to find D'bisect the arc of the segment. The rest of the construction is the same as before; and the triangle will evidently be isosceles.

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Prop. IX. PROB.—Given the three lines drawn from the vertex of a triangle, one of them perpendicular to the base, one bisecting the base, and one bisecting the vertical angle; to construct the triangle.*

Take any straight line BC and draw DA perpendicular to it, and equal to the given perpendicular; from A as centre, with radii equal to the lines bisecting the vertical angle and the base, describe arcs cutting BC

in E and F, and draw_AEH and AF; through F draw GFH perpendicular to BC, and to H, and cutting HG in G: from Gas centre, with GA as radius, describe a circle cutting BC in B and C : join AB, AC: ABC is the triangle required.

For (III. 3) since GFH is perpendicu. lar to BC, BC is bisected in F; and (III. 30) the arcs BH, HC are equal. Therefore (III. 27) the angles BAH, CAH are equal. Hence, in the triangle ABC, the perpendicular AD, the line AE bisecting the vertical angle, and the line AF bisecting the base, are equal to the given lines. Therefore ABO is the triangle required.

Schol. If the three given lines be equal, the problem is indeterminate; as any isosceles triangle, having its altitude equal to one of the given lines, will answer the conditions.

Method of Computation. Through A draw a parallel to BC, meeting HG produced in K. Then, in the right-angled triangle ADE, AE, AD being given, DAE, or H, may be computed; the double of which is A GK: and AK or FD being given, AG, GK can be found, and thence GF. Hence, if GB were drawn, it and GF being known, BF, and the angle BGF, or BAC, can be computed. The rest is easy; DAE, half the difference of B and C, being known.

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* Analysis. Let ABC be the required triangle, AD the perpendicular, and AE, AF the lines bisecting the vertical angle and the base. About ABC describe (IV.5) a circle, and join its centre G, with A and F, and produce GF to meet the circumference in H. Then (III. 3) GFH is perpendicular to BC, and (III. 30) the arcs BH, HC are equal. But (III. 26) the equal angles BAE, CAE at the circumference stand on equal arcs; and therefore AE being produced, will also pass through H: and the point H, and the angle GHA and its equal HAG are given. Hence also the centre G and the circle are given, and the method of solution is plain.

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Prop. X. PROB.—Given the base of a triangle, the vertical angle, and the straight line bisecting that angle; to construct the triangle. *

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On the given base BC describe (III. 33) the segment BAC capable of containing an angle equal to the given vertical angle, and complete the circle; bisect the arc BEC in E, and join EC; perpendicular to this, draw CF equal to half the line bisecting the vertical angle, and from F as centre, with FC as radius, describe the circle CGH, cutting the straight line passing through E and F in G and H; make ED equal to EG, and draw EDA; lastly, join AB, AC, and ABC is the required triangle.

For the triangles CEA, CED are equiangular, the angle CEA being common, and BCE, CAE being each equal to BAE. Therefore AE : EC :: EC : ED, and (VI. 17) AE.ED=EC?. But (III. 18 and 36) HE.EG or HE.ED=EC2; and therefore AE.ED=HE.ED; whence AE=HE, and (I. ax. 3) AD=GH=2CF. AD is therefore equal to the given bisecting line, and it bisects the angle BAC. Hence ABC is the required triangle.

Method of Computation. Draw EL perpendicular to BC, and join CH. Then BCE is equal to BAE, half the vertical angle A ; and therefore, to the radius EC; CL is the cosine of 1A, and CF is the tangent of CEF to the same radius: wherefore, to any radius, CL: CF, or BC : AD :: cosA : tan CEF or cot EFC"; and hence the angle H, being half of EFC, is known. Also ECH is the complement of H, because ECF is a right angle, and FCH equal to H. But (APP. III. 2) EC : EH or EX :: sin H : sin ECH, or cos H; or (APP. III. defs. cor, 6) EC:EA :: R:cot H. Also in the triangle ACE, EC : EA :: sin A : sinACE ; whence (V. 11) R : cot # :: sin A : sin ACE; whence ACE may be found; and if from it, and from ABE, its supplement (BE being supposed to be joined) BCE be taken, the remainders are the angles at the base.

* Analysis. Let ABC be the required triangle, and let AD, the line bisecting the vertical angle, be produced to meet the circumference of the circumscribed circle in E; join also EC. Then (III. 33) the circumscribed circle is given, since the base and vertical angle are given; and the arc BEC is given, as are also its half EC, and the chord EC. Now the triangles AEC, DEC are equiangular; for the angle CEA is common, and (III. 21)

BCE is equal to BAE or EAC. Hence AE: EC::EC: ED, and therefore AE.ED=EC2. Hence (III. 36) it is evident that if EC be made a tangent to a circle, and if through the extremity of the tangent a line be drawn cutting the circle, so that the part within the circle may ne equal to AD, DE will be equal to the external part: whence the construction is manifest.

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Prop. XI. PROB.— To draw a line parallel to a given straight line, such that the part of the parallel intercepted between two other straight lines given in position, may have a given ratio to the perpendicular distance between the two parallels.

Let AB be a given straight line falling on two other given straight lines AC, BC; it is required to draw a parallel to AB, such that the part of it between AC, BC may have a given ratio to the perpendicular distarice between AB and the parallel.

Through C draw CG perpendicular and CD parallel to AB, and take CD to CG in the given ratio; draw AED, and through E draw EF parallel to AB: EF is the line required.

Draw FH parallel to CG. Then, in the similar triangles, ACD, AFE, and ACG, AFH, CD : CA::FE : FA, and CA : CG ::FA: FH; whence, ex æquo, CD: CG :: FE : FH. FE, therefore, has to FH the given ratio, and it is parallel to AB.

Schol. 1. If DC be produced, and CD' be taken equal to CD, and if AD'E' be drawn, the straight line E'F' parallel to AB, would be proved in the same manner, to have the given ratio to its perpendicular distance from AB: it therefore equally answers the conditions of the problem. If, however, CD' be equal to AB, the second solution is inadmissible, as (I. 33) AD', BC would not meet. If CD' be greater than AB, E'È' will be on the other side of AB.

Schol. 2. If CD, CG, and consequently FE, FH, be equal, and if a parallel to FH were drawn through E, the figure EH would be a square; and we have thus the means of solving the problem in which it is required to inscribe a square in a given triangle, having one of its sides on an assigned side of the triangle. Now, since AB is to FE, and CG to CK, as CA to CF, we have AB : FE :: CG : CK; whence, alternately, and by inversion, CG : AB :: CK: FE, and, by composition, AB+CG : AB :: CK+FE : FE; that is, in case of the inscribed square, AB+ CG : AB :: CG: FE; and, therefore, the side of the inscribed square is a fourth proportional to the sum of the base and perpendícular, and to the base and perpendicular themselves.

It would be proved in a similar manner, that when CG, CD' are equal, E'F' would be a fourth proportional to the difference of the base and perpendicular, and to the base and perpendicular themselves.

Cor. BE' is divided harmonically (APP. I. 29) in C and E. For DD', a side of the triangle ADD', is bisected in C, and AB is drawn through A parallel to DD'. It would be shown in a similar manner, that AF' is divided harmonically in F and C.

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