M E C H 
the defcrlptlon of the curve, that the arc C D is equal to 
the line AD, and confequently the arc CH equal to 
D F or I K or C G ; but the arc C H is equal to the arc 
E G ; therefore C G is equal to the arc E G, and the or¬ 
dinate C K (= CG-fG K) mult be equal to the f’urn of 
the arc EG and the right line G K. 
Prop. LVL The line C H, parallel to the chord E G, is a 
tangent to the cycloid in C.—Draw an ordinate c k very near 
C K, meeting the curve in c, the circle in g, and the axis 
in k % let C u and G n, parallel to the axis, meet the or¬ 
dinate ck in u and n-, and from O, the centre of the 
circle EGF, draw the radius O G. Since c hz^E g+g h, 
therefore ct/^zGg+gn; and, if you iuppofe the ordi¬ 
nate ci to approach to the ordinate C K, and at length 
to coincide with it, as G g and Gn vanifli, the tri¬ 
angles Ggn and GOK become limilar; whence, Gif : 
gn :: OG : OK, and Gg-\-gn : gn :: OG-J-OK 
(=F K) : OK; but G« : gn :: G K : OK, therefore 
Gg+gn : Gn :: FK : GK :: GK : EK; and con¬ 
fequently cu : Cu :: GK : EK; and, if you draw the 
chord Cc, the triangles Cue, EKG, will be fimilar; fo 
that the chord Cc, as the points C and c coincide, be¬ 
comes parallel to E G: therefore the tangent of the cycloid 
at C is parallel to E G. 
Prop. LVII. The arc of the cycloid E L is double the chord 
EM of the correfponding arc of the generating circle E M F.— 
Let K L and k S be two very near ordinates of the cycloid, 
meeting the generating circle in M and Q; produce the 
chord EM till it meet the ordinate k S in P; let Q o be 
the perpendicular from Q on MP; then draw the lines 
EN and MN, touching the circle in E and M. Becaufe 
the triangles ENM, P Q M, are fimilar, and E N = N M, 
therefore P Q is equal to Q M ; and, the triangle PQM 
being ifofceles, the perpendicular Q o bifefts the bafe PM; 
fo that M P is double of M o : but, by the la(t propofi- 
tion, LS is parallel, and confequently equal, to M P, and 
LS is equal to zMo. The line LS is the increment of 
the curve EL, generated in the fame time that the chord 
E M increafes by M o, fince E Q is equal to E o, when the 
points Q and M corne together: therefore the curve in¬ 
creafes with double the velocity that the chord increafes; 
and, fince they begin at E to increafe together, the arc of 
the cycloid EL will be always double the chord EM. 
Cor. The femi-cycloid E L B is equal to twuce the dia¬ 
meter of the generating circle E F; and the whole cycloid 
A C E B is quadruple the diameter E F. 
Prop. LVI 1 I. Let ER be parallel to the bafe AB, and 
C R parallel to the axis of the cycloid ; and the [pace E C R, 
bounded by the arc of the cycloid EC and the lines ER and 
~RC,Jhall be equal to the circular area EGK.—Draw cr 
parallel to C R; and, fince cs : Ca :: GK : EK; there¬ 
fore EKx?»=GKxC#, and confequently RrXCR= 
GKxKf : therefore the little (pace CRrc— GKAg. So 
that the areas E C R, EGK, increafe by equal incre¬ 
ments; and, fince they begin to flow together, therefore 
they mult be equal. 
Cor. i. Let AT, perpendicular to the bafe A B, meet 
ER in T, and the fpace ETA CE will be equal to. the 
femicircle E G F. 
Cor. z. Since A F is equal to the femi-circumference 
EGF, the rectangle EFAT, being the reftangle of the 
diameter and femi-circumference, will be equal to four 
times the femicircle E G F: and therefore the area E CA FE 
will be equal to three times the area of the generating ie- 
micircle EGF. 
Cor. 3. If you draw the line E A, the area intercepted 
betwixt the cycloid EGA and the itraight line EA will 
be equal to the femicircle EGF; for the area EC AFE 
is equal to three times EGF, and the triangle EAF= 
AFx|EF the reftangle of the femircircle and radius, 
and confequently equal to 2EGF; therefore their diffe¬ 
rence, the area ECAF, is equal to EGF. 
Prop. LIX. Take EA=OK, draw AZ parallel to the bafe , 
meeting the generating circle in X, and the cycloid in Z, and 
join CZ, FX: then fiall the area CZEC be equal to the 
VOL. XIV. No. 1002, 
A N I C S. G49 
fum of the triangles GFK and b FX.—Draw Zd parallel 
to the axis E F, meeting E T produced in d, and the 
trapezium RCZ<A w ill be equal to £CR -EhZd X Rrfcr: (be¬ 
caufe Zaf=E^=0K) AOEX R d. But Rr/=RE-f E<f=CK + 
AZ=EG-{-GK-}-EX-f-AX; therefore the trapezium RCZ d 
is equal to the fum of the reftangles of half the radius 
and the arcs E G, E X, added to their fines G K and b X. 
But the area EGF, i. e. the triangle EGF and the feg- 
ment cut off by the chord E G, is equal to the reftangle 
contained by half the radius and the fum of the arc E G 
and its right fine GK; and the area E X F, confifting 
of the feclor E O X and the triangle X O F, is equal to 
the reftangle of half the radius and the film of the arc 
E X and its right fine l X ; therefore the trapezium R C Zd 
is equal to the fum of the areas EGF and E X F. By 
the laft Prop, the area EC R is equal to G K, and E'Zd 
—EbX. From the trapezium RCZ d f'ubtraft the areas 
E C R, E Zd ; and from the areas EGF, E X F, fubtraft: 
the areas EGK, EAX; and there will remain the area 
CZEC equal to the lum of the triangles GFK, AFX. 
Cor. 1. Hence, an infinite number of fegments of the 
cycloid may be afligned that are perfeftly quadrable. For 
example, if the ordinate C K be fuppofed to cut the axis 
iu the middle of the radius O E, then K and b coincide ; 
and the area ECK becomes in that cafe equal to the 
triangle GKF, and EAZ becomes equal to FAX; and 
thefe triangles themfelves become equal. 
Cor. z. Suppofe now that K comes to the centre O, and 
C comes tor; then, becaufe OK vaniflies, therefore EA 
vanifhes, and the fpace CZEC becomes in this cafe 
ECfE, which is equal to | 0 E 2 ; for the triangle AFX in 
this cafe vanifhes. 
Prop. LX. Let ATC (fig. 79.) be a femi cycloid, having 
its bafe E C parallel to the horizon, and its vertex A down¬ 
wards: fuppofe a firing , with a pendulum, of the length of the- 
femi-cycloid, fufpended at C, and applied to the femi cycloid 
CTA ; the body P, by its gravity, will gradually feparate the 
fring from the femi-cycloid CTA, and will deferibe an equal 
femi-cycloid A P V, having its vertex in V, and its axis per¬ 
pendicular to the horizon .—On the axis AE deferibe the 
generating femicircle AGE; draw A B cutting the ver¬ 
tical line CV in D; and on DV, taken equal to A E, 
deferibe the femicircle D H V. Then, fince the femi cy¬ 
cloid CTA is equal to zAE or CV, (by Cor. Prop. LVII.) 
therefore the body P will come to V when the firing CTP 
comes to a vertical fituation. Through T and P draw 
T G and PH parallel to A D, meeting the femicircles in 
G and H; and, fince the ftraight part of the firing TP 
is equal to the curve TA to which it was applied, there¬ 
fore TP~2AG=2TK, and confequently T K and KP 
are equal, and the points G and H mult be equally dif- 
tant tromthe line AD : and therefore the arc AG will be 
equal to D H, and confequently the angle GAD=;ADH; 
and the chords G A, D H, are parallel. But TP, being 
a tangent to the cycloid in T, is parallel to G A ; there¬ 
fore DKPH is a parallelogram, and D K is equal to PH. 
But the arc A G is equal to G T, by Prop. LV. and 
therefore the arc AGzrzAK ; and, fince AD=AGE, it 
follows that D K or P H=G E or H G : and, if P H be 
produced till it meet the axis in R, then (hall the ordi¬ 
nate P R be equal to the fum of the arc H V and its right 
fine HR; and therefore the point P, by Prop. LV. mull 
be in a femi-cycloid, whole generating circle is DHV, its 
axis D V, and vertex V. 
Cor. If another femi-cycloid equal to CTA, as Ct R, 
be placed in a contrary fituation, it is plain, that, by 
means of thefe femi-cycloids, a pendulum may be made 
to deferibe the cycloid AV B in its ofcilfations. 
Prop. LXI. Let V L, perpendicular to D V, be equal to any 
arc of the cycloid V M L ; deferibe with the radius V L the 
femicircle LZT, and, fuppofing the pendulum to begin an ofcil- 
lalion from L, the velocity acquired at M, in the cycloid, will 
be as M X the ordinate of the circle at the corresponding point 
M in the fir fight line V L ; and the force by which the motion 
of the pendulum is accelerated in M, is as the arc of the cy- 
S C cloid 
