surface. Because A and P are the poles of the circles BC and DE, the line AP is perpendicular to the planes of those circles (definition 2. Spher. Trig.); therefore AFB, AFC, PHD, PHE are right angles (def. 1. 2 Supp.); consequently FB is parallel to HD, and FC to HE (28. 1.); whence the angles BFC, DHE are equal (9. 2 Supp.) The lines BF, FC, DH, HE are also equal, being the sines of equal arcs; whence the arc BC is equal to DE (26. 3.)

Case. 2. When one of the planes passes through the centre of the globe, and the other does not.

Let ABDP (fig. 53.) be the circle formed by the intersection of the superficies of the sphere with the plane passing through the centre, ACEM the circle whose plane does not pass through the centre, AM the common section of the planes, O the centre of the circle ABDP; BF, DH the common sections of the planes BCF, DEH with the plane ABDP; join AO, MO, FC, LC, HE, NE, AC, ME; then, as in the first case, FB, FC, HD, HE are all equal; also AF MH. But the angle FAL = HMN (5. 1.), and AFL MHN, being both right; therefore AL- MN (26. 1.) Moreover, in the triangles ALC, EMN, the side AC = ME, and the angle CAL = EMN; hence LCNE (4. 1.): then, in the triangles LFC, EHN, we have LF NH, FC = HE, and

LCNE; therefore the angle LFC
BC= DE.

=

NHE, consequently

Case 3. When neither of the planes passes through the centre of the sphere.

Through the common section of these planes and the centre of the sphere, let a third plane pass; then, by the last

case, the arches of one of the equidistant circles, intercepted between the third plane and each of the others, and the corresponding arcs of the other similarly intercepted arc, are equal, each to each; and therefore their sums or differences will be equal. But the sum of those arcs, when the given cutting planes fall on the opposite sides of the third plane, and their difference when they fall on the same side, will be the arcs intercepted between the first mentioned cutting planes.

X
PROP. XI.

If EFGH, efgh (fig. 51. and 54.) represent two equal circles, whereof EFGH is as far from its pole P, as efgh is from the projecting point; any two right lines EP, FP drawn through P will intercept the representatives of equal arcs of those circles; on the same side, if P falls within the circles; but on the contrary side, if without; that is EF = ef, and GH =gh.

For, by the Lemma, two planes passing through the pole of the original circle, represented by EFGH, and the projecting point, will cut off equal arcs of those circles, and these planes will intersect the plane of projection in right lines passing through the projected pole P.

COROLLARY I.

If a circle is projected into a right line EF (fig. 55.) perpendicular to the line of measures EG; and if from the centre of the primitive C, a circle ef P be described, passing through

P the pole of EF, and PF be drawn; then the arc ef= EF. And if any other circle be described whose vertex is P, the arc ef will always be equal to EF.

COROLLARY II.

If from the projected pole of a great circle there be drawn two right lines, the intercepted arc of that circle will be equal to the intercepted arc of the primitive.

COROLLARY III.

If from an angular point, two right lines be drawn through the poles of its sides, the intercepted arc of the primitive will be the measure of that angle.

For the arc of a great circle contained between the poles measures the inclination of the axes, or the inclination of the planes, of those circles.

PROP. XII.

If QH, NK (fig. 56. and 57.) be two equal circles, whereof NK is as far from the projecting point as QH from its pole P; and if they be projected into the circles whose radii are MC or CL, and DF or FG, F being the centre of DG, and E the projected pole; I say, the pole E will be distant from their centres in proportion to the radii of the circles; that is, CE: EF:: CL: DF or FG.

For, since NK and ML are parallel, and the arch NI= PH, therefore angle ELINKI (or n KI) = GIP; there

fore the triangles IEL and IEG are similar; whence EL: EI:: EI: EG.

Again, the angle EMI KNI= PIQ, and therefore the triangles IEM and IED are similar; whence EM: EI:: EI: ED. Therefore EI2 = EL × EG=EM × ED; consequently EM: EL::EG: ED; and, by composition, EMEL EM-EL EGED EG-ED

; that

2

COROLLARY I.

Hence, if the circle KN be as far from the projecting point as QH is from either of its poles, and if E, O be its projected poles; then will EL: EM::ED: EG:: OD: OG.

This follows from the foregoing demonstration and prop. 9.

COROLLARY II.

Hence, also, if F be the centre, and FD the radius of any circle QH, and E, O the projected poles; then EF: DF:: DF: FO.

COROLLARY III.

Hence if the circle DBG (fig. 58.) be as far from its projected pole P, as LMN is from the projecting point; and if any right lines be drawn through P, as MPG, NPK, they will cut off similar arches GK, MN in the two circles.

For, from the centres C, F, draw the lines CN, FK; then, since the angles CPN and FPK are equal, and by this prop. CP: CN: FP: FK; therefore (by 7. 6.) the triangles. PCN and PFK are similar; and the angle PCN = angle PFK; therefore the arches MN and GK are similar.

COROLLARY IV.

Hence, also, if through the projected pole P of any circle DBG, a right line BPK be drawn; then, I say, the degrees in the arc GK shall be the measure of DB in the projection ; and the degrees in DB, shall be the measure of GK in the projection.

For (by prop. 11.) the arc MN is the measure of DB, and therefore GK, which is similar to MN, will also be the measure of it.

COROLLARY V.

The centres of all projected circles are beyond the projected poles (in respect to the centre of the primitive); and none of their centres can fall between them.