TRICON 
toned about bc, so as to pass through the point a ; the right 
lines ba, ca, having respectively the two points B and a, c 
and a, in this plane, must themselves lie in it. Hence, the 
three right lines bc, ba, ca, being in the same plane, the 
three right lines mn, op, qr, would likewise be in it, by the 
preceding, q.e.d. 
(3.) Let abcd, efgh, (fig. 21.) be two planescuttingone 
another in the line mn. Then, this line Mn is a right one. 
Suppose the line mn, which is common to the two planes, 
not to be a right line. In this case the points M and n might 
be joined by a right line min in the plane abcd; and, as 
this right line is supposed different from mn, it is not in the 
plane efgh, and there might be another right line, mow, 
drawn in the plane efgh, between the same points m and n. 
Consequently, on this supposition, the two right lines min, 
mon, would enclose a space, which is impossible. Hence, 
the supposition is false; that is, the line of intersection mn 
is a right one. q.e.d. 
(4.) Let ab (fig. 22.) be a right line perpendicular to both 
the right lines gh, ik, at their point of intersection, b. 
Then, ab is perpendicular to the plane cdef, in which these 
lines gh, ik, lie. 
Through the point b draw any right line EM, in the plane 
cdef. Take the portions bg, bi, bh, bk, equal to each 
other; and draw the right lines ag, ai, ah, ak. In the 
four triangles thus formed, abg, abi, abh, abk, the four 
angles abg, abi, abh, abk, are granted equal, being all 
right angles. Likewise the side ab is common, and there¬ 
fore as the sides bg, bi, bh, bk, have been taken equal, the 
four bases ag, ai, ah, ak, are equal by Geom. 99; and as the 
sides bi, bg, are equal respectively to the sides bh, bk, and 
the contained angles ibg, hbk, being vertically opposite, are 
also equal, therefore the angles bil, bkm, are equal; and 
also the bases ig, hk. Consequently, the three sides ag, 
ai, ig, of the triangle aig having been proved respectively 
equal to the three sides ah, ak, hk, of the triangle ahk, 
the angle ail opposite ag is equal to the corresponding 
angle arm opposite ah. But in the triangles ibl, kbm, the 
vertically opposite angles ibl, kbm, are equal; as also the 
angles bil, bkm ; and the sides bi, bk: therefore, by Geom. 
99, il=km, and bl = bm. Thus, in the triangles ail, 
akm, we have shown that ai = ak, il = km, and the angle 
ail —akm; wherefore, by Geom. 99, al=am. Hence, 
as in the triangles abl, abm, it has been proved that al = 
am, and bl = bm, the side ab being common, we have the 
angle abl = abm, by 101 Geom.; that is, ab is perpen¬ 
dicular to lm. In the same manner it may be demonstrated 
that ab is perpendicular to every other right line in the plane 
cdef ; whence, by Def. I. ab is perpendicular to the plane 
itself, q.e.d. 
(5.) Let adbea, (fig. 23.) be a plane section of a sphere. 
Then, adbea is a circle. 
Dem. First : If the plane pass through the centre, c, of 
the sphere as in figure 23. Drawjright lines, cd, ce, cb, &c., 
from the centre to all points, d, e, b, &c., of the boundary 
of the section ; and as all these right lines are drawn from 
the centre to the surface of the sphere, they are all equal, by 
Def. VII. Then abdea is a circle. 
Secondly : If the plane do not pass through the centre, 
c, of the sphere, as in figure 24. Draw the right line cf per¬ 
pendicular to the plane of the section adbea; and also a 
right line, cd, ce, to any points, d, e, &c„ of the boundary 
of the section. Join fd and fe. By Def. II. cf is perpen¬ 
dicular to fd and fe; therefore, by 111 Geom., cd 2 — cf 2 
4- fd 2 , and also ce 2 — cf 5 -f- fe 2 . Consequently, as cd — 
ce by Def. VI, cf 2 + fd 2 — cf 2 4- fe 2 , .*. fd 2 rr fe 2 , fd 
— fe. Hence, all the right lines, such as fd, fe, &c., 
drawn from the point f to the boundary of the section, being 
equal, this boundary is a circle, q.e.d. 
(6.) Because their respective radii, being all radii of the 
same sphere, are equal. 
(7.) The intersection of the planes of two great circles is 
a diameter of the sphere, and a common diameter of both 
circles; it therefore divides both into equal parts. 
O M E T R Y. 99 
(8.) Let abd, ghi, (fig. 25.) be two small circles equally 
distant from the centre c (that is, having the perpendiculars 
cf, ce, drawn to them from the central, equal). Then, the 
circle abd, ,ghi, are equal to each other. 
Dem. Draw cb, cg, from the centre c, to any points b, 
g, in the circumferences abd, ghi ; and join fb, eg. Con¬ 
sequently, as cf, ce, are, by Def. I. perpendicular to fb 2 
eg, we have, by Geom. Ill, cb 2 = cf 2 + fb 2 , and also cg, 
~ ce 2 eg 2 . But cb — cg, being radii of the sphere, 
cf 2 4- fb 2 — ce 2 4 - eg 2 , or as cf — ce fb — eg. Hence, 
as the radii of the two circles abd, ghi, are equal, the cir¬ 
cles themselves must be equal, q.e.d. 
(9.) Let abc be a triangle, formed on the surface of a 
sphere whose centre is o, by the planes abo, aco, cbo, of 
three great circles, whose arcs ab, ac, cb, form the three 
sides of the triangle. Also, let the planes abo, aco, be 
perpendicular, that is, let the angle bac of the triangle be a 
right one. Then, calling the three angles, bag, abc, acb, of 
the triangle (that is, the three angles under the planes abo 
and aco, abo and cbo, aco and cbo), a, b, c, respectively, 
and the sides bc, ac, ab, opposite these angles respectively, 
a, b, c ; we shall have 
sin a X sin b — sin b, and sin a x sin c — sin c. 
Dem. From the vertex of the angle c draw the right line 
cd perpendicular to oa ; and from the point d draw the 
right line de perpendicular to ob. Join ec by a right line. 
Now, cd being perpendicular to oa, is perpendicular to the 
plane abo ; because drawing in this plane the right line df 
at right angles to oa, cd will then be perpendicular to do 
by construction, and to df, by Def. III., inasmuch as the 
planes abo, ago, are perpendicular to each other: conse¬ 
quently (by 1.) cd is perpendicular to the plane abo. But 
cd being perpendicular to the plane abo must be also per¬ 
pendicular to the line de, by Def. I.; and therefore cde is 
a right-angled triangle. Hence, in the right-angled triangles 
cdo, deo, cde, we have, (by the 47 of the 1st of Euclid.) 
DC 2 — OD 2 4- CD 2 ; 
and od 2 — oe 2 4- de 2 ; 
and ec 2 — ed 2 -j- cd 2 , ec 2 — ed 2 — cd 2 . 
Putting the values for od 2 and cd 2 found from the last two 
equations into the first, we get 
oc 2 — oe 2 + ec 2 ; 
consequently, by the triangle oec is also right-angled at e. 
Thus, dk and ce being drawn respectively in the planes abo, 
cbo, perpendicular to ob, the intersection, we have the anlge 
dec equal to the angle between the planes, by Def. II. that is 
to the angle b. Likewise, in the right-angled triangle cde, 
CD _ CD 
sin dec — sin B — —. But, — — sin doc — sm (of 
corresponding arch) ac — sin b, .% cd — co x sin b; also 
CE 
— — sin coe — sin (of corresponding arch) bc — sin a, 
CO 
ce — co x sin a. Hence, sin b — — —.9? * sin ^ 
CE co X sm a 
sin a X sin b — sin b. 
Instead of beginning with the vertex of angle c, we might 
as well have begun from that of angle b, and the process 
would have been exactly similar, only that for angle b we 
should have had angle c, and for b the side opposite b, we 
should have had c the side opposite c. Making these sub ¬ 
stitutions in the above formula, we obtain, without going 
through another demonstration, sin a x sin c. — sin c. 
Q. E. D. 
(10.) In the preceding figure, representing the sides and 
angles as before, we shall have 
tan c — tan a x cos b, and tan b — tan a X cos c. 
Dem. In the right-angled triangle cde, cos dec, or cos b, 
— —But, — — tan doe — tan (of arch) ba — tan c, 
CE OE J 
CE 
DE — OE X tan c; also, —( ” tan coe — tan (of arch) 
bc — 
