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CAd and the common angle D.A.d, therefore the triangles B.A.d, DAC

are similar.

Corollary. Let ABC be any triangle, and let straight lines AD, BD, CD be drawn to D, any point in its plane; at the point B in the line BA make the angle ABd equal to the angle ADC, viz. that which the side opposite to B subtends at D; at the point C in the line CA make the angle ACd equal to the angle ADB, which AB the side opposite to C subtends at D; draw a line from the remaining angle A to d the intersection of the lines Bd, Cd; the triangles ADC, ABd are similar; and the triangles ADB, ACd are similar.

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5. The truth of the Corollary may be inferred from the theorem: it may however be proved directly as follows.

Let E be the intersection of the lines dB, CD (fig. 1.); join AE. Because by construction the angles ADC, ABd are equal, the angles ADE, ABE are equal; therefore the points A, D, B, E are in the circumference of a circle; hence the angle AEB is either equal to the angle ADB, or is its supplement; now by hypothesis the angle ADB is equal to ACd; therefore AEB or A.Ed is either equal to ACd, or is its supplement; hence, in each case, the points A, d, C, E are in the circumference of a circle; and therefore the angle ACD or ACE is equal to Ad B; now, by construction, the angle ADC is equal to the angle ABd; therefore the triangles ADC, ABd are similar; and since the angle CAD is equal to. dAB; by adding or subtracting the angle dAD, we have the angle CAd equal to DAB; now the angle ACd is by construction equal to ADB ; therefore the triangles ACd, ADB are similar.

6. In the demonstration, it was assumed that when the angles BAD, CAd are equal, then B.A.d, CAD are equal; this will always be true when the lines AD, Ad are either both within the angle BAC, or both without that angle, or, which is the same thing, when the similar triangles BAD, dAC are similarly situated; and the same is true of the similar triangles B.A.d, DAC.

7. I shall now apply the geometrical theorem to the construction of the problem enunciated in the beginning of this paper.

PROBLEM. (Figs. 1, 2, 3.)

Three stations A, B, C are given in position; also, there are given the angles ADB, ADC which the lines joining A, one of them, and B, C, the other two, subtend at a fourth station D in their plane; to determine the position of the fourth station, by a geometrical construction.

Solution. At the point B, in the line BA, make the angle ABd equal to ADC the angle which AC subtends at D; observing, that the line Bd must have such a position, that if the angle ABd were placed on ADC, so that BA lay along DA; then Bd would lie on DC. Also, at the point C, in the line CA, make the angle ACd equal to ADB, the angle which AB subtends at D; observing that the position of Cd must be such as to admit of the angle ACd being applied on ADB ; join A, and d the intersection of the lines Bd, Cd. By the theorem, the triangles ADC, ADB will be similar to ABd, ACd respectively; and as all the angles of these last are manifestly given, because their sides are given in position, therefore all the angles of the triangles ADC, ADB will be known; the angle DCA being equal to Bd A; DBA to Cd A; DAC to BAd; and DAB to CAd.

Scholium. Since the angle Bd A is equal to DCA (fig. 2.) and Cd. A to DBA, the angle Bd6' is equal to the sum of the angles DCA, DBA. Now it may happen that their sum is equal to two right angles; then, Bd and Cd will be in one straight line; and, there being no intersection, nothing can be determined in respect to the angles BAD, CAD. But in this case, since DCA, DBA make two right angles, the points A, B, C, D are in the circumference of a circle. Thus it appears, that when the point D is in the circumference of a circle which passes through A, B, C, the problem is indeterminate.

8. It is deserving of remark that this simple construction, by which the point d is found, has served to change the proposed geodetical problem into another which at first view appears easier of solution; for since the angle ABd=ADC is given (figs. 1, 2, 3.), and also the angle ACd = ADB ; and moreover, because the angles which the lines AB, AC make with a line drawn from B to C may be considered as known, the angles dBC, dCB are known.

The proposed problem is now transformed to this:

“Having given all the sides, or else all the angles of two triangles ABC, dBC which have a common base BC, it is required to find the angles which the line Ad joining their vertices makes with the sides.”

This is the geometrical expression of a well known geodetical problem which is more frequently resolved than the other; probably, because its solution is supposed easier. The geometrical property of the figure by which the one problem is converted into the other, namely, the equality of the rectangles AD. Ad and AB.AC is easily remembered, a circumstance of considerable importance in practical applications of Geometry.

9. The following Theorem is a deduction from the proposition:

“If straight lines B.D. BE be drawn from B, one of the angles of a triangle ABC, making equal angles ABE, CBD with the sides about

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that angle; and also straight lines CD, CE, making equal angles BCD, ACE with the sides about another of the angles, and meeting the former lines in D, E, then, straight lines drawn from the remaining

angle 4 to D and E shall make equal angles BAD, CAE with the

sides about the angle”.” The well known proposition, that the lines which bisect the three

angles of a triangle meet in the same point is a particular case of this

Theorem.

10. The Trigonometrical solution of the Geodetical Problem of Art. 1. which is deducible from the construction here given, is sufficiently obvious: I shall therefore, without at present entering into it, investigate another theorem which comprehends the former and various others.

Let the sides of any triangle be a, b, c, (fig. 5, 6).
The opposite angles

Fig. 5.

Let straight lines be drawn from any point D to the angles of the triangle, and put AD=a, BD = y, CD=x.

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* I owe this elegant proposition to T. Galloway, Esq. F. R.S. to whom it occurred when considering the Theorem. Vol. VI. PART I. P

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where it must be observed that the angles a, 3, y, must be so reckoned, that their sum is four right angles.

By a known Trigonometrical formula,

a’– 24 y cos y + y = c .................. (1),
a”-24'x cos B+ 2* = b .................. (2),
3/*— 23/x cos a + x*= a .................. (3).

The condition that the triangle ABC is made up of the three triangles ADB, ADC, BDC (fig. 5), or else of the excess of two of

them above the third (fig. 6), is expressed analytically by this other equation,

a y sin y + æ & sin (3 + yx sin a = b c sin A ......... (4).

These hold true, whatever be the position of the point D, observing

always that the angles a, B, y must satisfy the condition of making up four right angles.

By adding the first and second equations, and subtracting the third, we obtain,

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Let equation (5) be multiplied by sin a, and equation (4) by cos a : the results are

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By subtracting the second of these equations from the first, and observing that

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