The two triangles CET, CAN have then the angle c common, and the sides about that angle reciprocally proportional; those triangles are therefore equal, namely, the ▲ CET = A CAN. Corol. 1. From each of the equal tri. CET, CAN,

take the common space CAPE,

and there remains the external ▲ PAT=▲ PNE.

Corol. 2. Also from the equal triangles CET, CAN, take the common triangle

CED,

and there remains the ▲ TED trapez. ANED.

THEOREM XIX (19).

The same being supposed as in the last Proposition; then any Lines KQ, QG, drawn parallel to the two Tangents, shall also cut off equal Spaces. That is,

AKQG trapez.ANHG,

and AKqg=trapez. ANhg.

H

G C

Corol. 2. Hence also, CA2 = CD2 + cd2,

and ca2 = DE2 + de2.

Corol. 3. Further, because Ca2: ca:: AD.DBor cd: DE2,

therefore CA : ca :: cd : DE.
likewise CA : ca :: CD : de.

THEOREM XVII (15).

If from any Point in the Curve there be drawn an Ordinate, and a Perpendicular to the Curve, or to the Tangent at that Point: Then, the

Dist. on the Trans, between the Centre and Ordinate, CD:

For, by theor. 2, CA2: Ca2 :: AD.DB : DE,
But, by rt. angled As, the rect. TD.DP = DE2;

and, by cor. 1, theor. 16,
therefore

CD. DT = AD.DB;

- CA2: Ca2 :: TD.DC : TD.DP,

AC: Ca2:: DC

THEOREM XVIII (18).

If there be Two Tangents drawn, the One to the Extremity of the Transverse, and the other to the Extremity of any other Diameter, each meeting the other's Diameter produced; the two Tangential Triangles so formed, will be equal.

For, draw the ordinate DE. Then

By sim. triangles, CD : CA :: CE : CN;
but, by theor. 7,

CD: CA:: CA: CT;

theref. by equal. CA : CT:: CE : CN.

The

The two triangles CET, CAN have then the angle c common, and the sides about that angle reciprocally proportional; those triangles are therefore equal, namely, the A CET = A CAN. Corol. 1. From each of the equal tri. CET, CAN, take the common space CAPE,

and there remains the external A PAT = A ΡΝΕ.

Corol. 2. Also from the equal triangles CET, CAN,

take the common triangle
and there remains the

CED.

TED = trapez. ANED.

THEOREM XIX (19).

The same being supposed as in the last Proposition; then any Lines KQ, QG, drawn parallel to the two Tangents, shall also cut off equal Spaces. That is,

are to each other as CA2, CD, CG2;

The three sim. triangles CAN, CDE, CGH,

th. by div. the trap. ANED: trap. ANHG :: CA2 - CD2: CA2 - CG2.

GQ2::CA-CD2: CA2 - CG2.

theref. by equality, ANED :

But, by cor. 2, theor. 18, the trap. ANED = A TED;

and therefore the trap, ANHG = AKQG.

In like manner the trap. ANhg = AKqg. Q.E.D. Corol. 1. The three spaces ANHG, TEHG, KQG are all equal. Corol. 2. From the equals ANHG, KQG,

take the equals Anhg, Kqg,
and there remains ghHG = gqQG.

Corol. 3. And from the equals ghao, gqag,
take the common space gqLHG,

and there remains the ALQH = A Lqh.

Corol. 4. Again from the equals KQG, TEHG, take the common space KLHG,

the triangle

KQG becomes the triangle IRC,

and the space ANHG becomes the triangle ANC ;
and therefore the A IRC =

ANC = A TEC.

Corol. 6. Also when the lines ka and HQ, by moving

with a parallel motion, come into the position ce, Me,
the triangle LQH becomes the triangle ceм,
and the space TELK becomes the triangle TEC;

and theref. the A cem = A TEC = A ANC

IRC.

THEOREM XX (20).

Any Diameter bisects all its Double Ordinates, or the Lines drawn Parallel to the Tangent at its Vertex, or to its Con

jugate Diameter.

For, draw an, qh perpendicular to the transverse. Then by cor. 3 theor. 19, the A LQH = A Lqh; but these triangles are also equiangular;

consequently their like sides are equal, or LQ=Lq.

Corol. Any diameter divides the ellipse into two equal parts.

For, the ordinates on each side being equal to each other, and equal in number; all the ordinates, or the area, on one side of the diameter, is equal to all the ordinates, or the area, on the other side of it.

THEOREM

THEOREM XXI (21).

As the Square of any Diameter :
Is to the Square of its Conjugate ::
So is the Rectangle of any two Abscisses :

To the Square of their Ordinate.

That is, CE: Ce2 :: EL. LG or CE2

For, draw the tangent TE, and produce the ordinate QL to the transverse at K. Also draw

ан, ем perpendicular to the transverse, and meeting EG in H and M. Then, similar triangles

TKA

being as the squares of their like sides, it is,

CL2: LQ2.

Q

e

H

B

C

by sim. triangles, A CET: A CLK :: CE : CL; or, by division,

ACET: trap. TELK :: CE2: CE2

Again, by sim. tri. A cem : ALQH :: ce2 : LQ2.

But, by cor. 5 theor. 19, the A cem = A СЕТ,
and, by cor. 4 theor. 19, the A LQH = trap. TELK;
theref. by equality, CE : ce :: CE2

or

CL2: LQ,

Q. E. D.

CE2: Ce :: EL.LG: LQ2. Corol. 1. The squares of the ordinates to any diameter, are to one another as the rectangles of their respective abscisses, or as the difference of the squares of the semidiameter and of the distance between the ordinate and centre. For they are all in the same ratio of ce2 to ce2.

Corol. 2. The above being the same property as that belonging to the two axes, all the other properties before laid down, for the axes, may be understood of any two conjugate diameters whatever, using only the oblique ordinates of these diameters, instead of the perpendicular ordinates of the axes; namely, all the properties in theorems 6, 7, 8, 14, 15, 16, 18 and 19.

THEOREM XXII (22).

If any Two Lines, that any where intersect each other, meet the Curve each in Two Points; then

The Rectangle of the Segments of the one :
Is to the Rectangle of the Segments of the other ::
As the Square of the Diam. Parallel to the former :
- To the Square of the Diam. Parallel to the latter.

That