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2. In comparatively modern times an interesting application of the problem was made to geodetical measurements. In the year 1620 Snellius, when ascertaining the distance between Bergen-op-zoom and Alcmaer, with a view to the determination of the magnitude of the Earth, employed it in finding the position of his Observatory. He assumed as given points three stations whose positions had been determined, and taking the angle which each two of them made at the Observatory, he was able to determine, by a trigonometrical computation, the distances of the stations, and thence its position*. The same problem was proposed by Richard Towneley as a chorographical problem†, and resolved trigonometrically in the Philosophical Transactions about the year 1670, by John Collins.

3. We are informed by Delambret that Lalande wishing to compute some observations of the Moon which had been made at the Military School, Paris, proposed to find the longitude of the station where the observations had been made, by observing there the angles subtended by three steeples whose positions were known. He was thus led to the same application as had been made long before by Snellius, without knowing or without thinking of his solution. Lalande's patience was exhausted by the length of the calculations, and the slips he made in performing them: he therefore referred the problem to Delambre, who gave a solution which was printed in Cagnoli's Trigonometry (First Edition), and again, but with more detail, in his own treatise Methodes Analytiques pour la Determination d'un Arc du Meridien.

Delambre's solution, which is analytical, is good, his formulæ have however but little of that symmetry and simplicity which constitute elegance in a geometrical speculation, and make it easy to be comprehended and remembered.

4. In considering the problem I have found two Theorems; from one of them a particularly simple and elegant geometrical construction

* SNELLIUS, Erastosthenes Batavus, p. 203.

+ Lowthorpe's Abridgement of Phil. Trans. Vol. I. p. 120.
Histoire de l'Astronomie Moderne, T. II. p. 109.

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is obtained, and both suggest various solutions, some lineo-angular, others trigonometrical, to this and other related problems.

THEOREM.

"Let AB, AC be two straight lines which meet in A; and AD, Ad other two, which make with the former equal angles BAD, CAd, these last being either both within (figs. 1, 2.) or both without the angle

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BAC (fig. 3).
Let the lines be such that the rectangle DA.Ad is
equal to the rectangle BA.AC; draw lines from D and d to B and C:
the triangles ADC, ABd, thus formed, are similar; and the triangles
ADB, ACd are similar."

DEMONSTRATION. Because by hypothesis AD. Ad=AB. AC, therefore AB AD=Ad : AC. Now the angles BAD, dAC are equal by hypothesis, therefore the triangles BAD, dAC are similar.

Also because AB : Ad=AD : AC, and the angles BAd, DAC are equal, for they are the sums or differences of the equal angles BAD,

CAD and the common angle DAd, therefore the triangles BAd, 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.

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 AEd 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 AdB; 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 BAd, 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 BAd, DAC.

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

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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 con

struction.

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 4, 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 BdA; DBA to CdA; DAC to BAd; and DAB to CAd.

Scholium. Since the angle BdA is equal to DCA (fig. 2.) and CdA to DBA, the angle BdC 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 BD.BE be drawn from B, one of the angles of a triangle ABC, making equal angles ABE, CBD with the sides about

Fig. 4.

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D

B

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

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