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BG' are, each of them, in the same plane as CA, and

not parallel to it, they will both meet it: let them meet

[blocks in formation]

CA, produced, in G and G', respectively; then, since (E. 28.3.) the arch ABC is equal to ADC, and that AB is equal to AD, therefore, also, the arch BC is equal to DC: whence, (E. 29. 3.) the chord AB is equal to the chord AD; the angle BAC is equal (E. 27. 3.) to DAC, and the angle BCA to DCA; therefore, (E. 13. 1.) the angle BAG' is equal to DAG; and the angle G'BA, being equal (E. 32. 3.) to BCA, and the angle GDA to DCA, therefore the angle G'BA is equal to GDA; and the chord AB has been shewn to be equal to AD: therefore, (26. 1.) the two triangles BAG' and DAG are equal to one another. So that BG' is equal to DG, and AG' to AG; that is, the tangents, drawn from B and D, are equal to one another, and meet CA produced in the same point.

And, in the same manner, may the proposition be proved, if the two circles meet only in one point, A; that is, if they touch one another: for then (Art. 147.) the common section of their planes, AG, will touch them both; and it follows, from E. 36. 3. that G'B is equal to G'A, and GD to GA; wherefore, (E. 5. 1.) the angle G'AB is equal to G'BA, and the angle GAD to GDA: but the angle G'BA may be shewn, as before, to be equal to GDA; and AB to be equal to AD: therefore, AG' is equal to AG, and the two tangents meet in the same point G, of the common section of the circles: and, therefore, each being equal to the same straight line GA, they are equal to one another.

PROP. XV.

(170.) Theorem. If two circles in a sphere meet each other, and two straight lines be drawn, the one touching the one circle, and the other the other, the two tangents so drawn, if they be parallel, are each of them parallel to the common section of the circles, and each of them is drawn through the bisection of the arch which it touches and if they meet, they meet in the same point of the common section of the circles, and are equal to one another.

If the two tangents be parallel, and one of them also be parallel to the common section of the circles, the other must (E. 9. 11.) be likewise parallel to that section: But if it be possible, let neither of them be parallel to the common section; therefore, since each tangent is in the same plane with that section, they must both, if produced meet it; and consequently, (E. 7. 11.) the

two tangents, and the common section of the circles, are all in the same plane; that is, the two circles are in the same plane which is contrary to the supposition. Wherefore, the tangents are each of them parallel to the common section; and it is evident, from E. 29. 1. 3. 3. and 30. 3. that they pass through the bisections of the arches which they touch.

B

E

H

But, let ABC, ADC, be circles, which meet each other, in a sphere, having GC for their common section; and let the tangents BG, DG, meet each other in G. Then BG and DG, being each in the same plane with CA, must both meet it; and since the three straight lines BG, DG and CG are not all in the same plane, they can only meet in one point; otherwise they would have three points of intersection, and therefore, (E. 2. 11.) they would all be in the same plane; which is contrary

to the supposition. Therefore, BG, DG, and CG all meet in the same point G: and since (E. 36. 3.) each of the squares of GB and of GD, is equal to the same rectangle CG, GA, if the circles cut each other, or to the same square of GA, if they touch one another in A, therefore, GB is equal to GD.

(171.) COR. 1. If the two circles ABC, ADC be equal, and the two tangents BG, DG meet each other, then the arch AB is equal to the arch AD.

For, if AB be not equal to AD, make (E. 1. 4. and 28. 3.) the arch AH equal to AD; and at H suppose a tangent, HG',,to be drawn to the circle ABC: wherefore, (Art. 160.). HG' meets DG, in G, and is equal to it but GD has been proved to be equal to GB: therefore, GH is equal to GB, which (E. 8. 3.) is absurd; for, it is plain, that another tangent, GB', may be drawn to the circle ABC, on the other side of GA, which tangent (E. 36. 3.) is equal to GB.

(172.) COR. 2. From the demonstration, it is manifest, that if each of any three straight lines, which are not all in the same plane, meets both the others, the three straight lines must all meet in the same point.

PROP. XVI.

(173.) Problem.

In the arch of a great circle, join

ing the intersections of two given lesser circles, in a sphere, to find a point, from which, if an arch of a great circle be drawn so as to touch either of the given circles, it shall be a quadrant.

Let Bx be an arch of a great circle joining the two points of intersection A and B, of two given lesser circles in a sphere, ABC, ABD. It is required to find a point

B

D

in Bx, from which an arch of a great circle, being drawn so as to touch either ABC, or ABD, it shall be equal to a quadrant.

Find (Art. 64.) the poles E and Fof ABC and ABD; join E, F (Art. 66.); and produce EF, both ways, to meet the circumferences of the given circles in C and D; from either of the two last mentioned points, as C, describe (Art. 70.) the arch of a great circle CI, at right angles to CED, and let CI meet Bx in I: I is the point required.

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