equal to AB (part 1, of this), either of them, as FM, is equal to its half CB; in like manner FN may be proved equal to CB; therefore FM and FN are equal, and of course the angles FMN and FNM (5. 1 Eu.), whence, the triangles MCF and NCF having also the right angles at C equal, MC is equal to CN (26 1. Eu.), and so MN is bisected in C..

In fig. 2, MB and BN, being drawn, are equal (Def. 5. 1 Sup.), and therefore the angles BMN and BNM (5. 1 Eu.); whence, the triangles MCB and NCB, having also the right angles at C equal, MC is equal to CN (26. 1 Eu.), and so MN is bisected in C.

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Cor. "The distance of a vertex (M, see fig. 1), of the second "axis (MN) of an ellipse from either focus (E or F), is equal to the principal semiaxis." It having been proved, in the demonstration of part 3 of this proposition, that either EM or MF is equal to CB.

PROP. II. THEOR.

The square of the second semiaxis (CM, see fig. to prec. prop.), of an ellipse or hyperbola (BP), is equal to the difference of the squares of the principal semiaxis (CB), and the eccentricity (CF), or to the rectangle under the distances of either focus (as F), from the principal vertices (A and B), or of either principal vertex (as B), from the focuses (E and F).

In the ellipse (see fig. 1), draw MF, which is equal to CB Cor. 1. 1 Sup.]; whence, in the triangle CFM, right-angled at C, the square of CM is equal to the difference of the squares of MF and CF [47. 1 Eu.], or of CB and CF, or, which is equal [Schol. 6. 2 Eu.], to the rectangle AFB or EBF.

In the hyperbola (see fig. 2), draw MB, which is equal to CF [Def. 5. 1. Sup.]; whence, in the triangle CBM, right angled at C, the square of CM is equal to the difference of the squares of MB and CB [47. 1 Eu.], or of CF and CB, or, which is equal [Schol. 6. 2 Eu, to the rectangle AFB or EBF.

Cor 1. Hence, if, in ellipses, there be taken two points (E and F, see fig. 1), in the principal axis (AB), the rectangle under the distances of each of which from the principal vertices (A and B, namely, the rectangle AEB or AFB), is equal to the square of the second semiaxis (CM), which may be done by cor. 2. 5 and 6. 2 Eu. these points are the focuses of the ellipse. Cor. 2. In like manner, if, in hyperbolas, there be taken two points (E and F, see fig. 2), in the principal axis (AB) pro

duced both ways, the rectangle under the distances of each of which from the principal vertices (A and B, namely, the rectangle AEB or AFB), is equal to the square of the second semiaxis (CM), which may be done by còr. 3. 5 and 6. 2 Eu these points are the focuses of the hyperbola or opposite hyperbolas.

PROP. II. THEOR.

The sum of the distances (GE and GF, see fig. 1), of any point (G) without an ellipse (APB), from the focuses (E and F), is greater, of any point (H) within it, less, than the principal axis (AB).

And the difference of the distances (GE and GF, see fig. 2), of any point (G) without opposite hyperbolas (AO and BP), from the focuses (E and F), is less, of any point (H), within one of them (BP), greater, than the principal axis (AB).

And the distance (GF, see fig. 3), of any point (G) without a parabola (KP), from the focus (F), is greater, of any point *(H) within it, less, than the distance of the same point from the directrix (DO).

Part 1, fig 1. Let FG, or FH produced, meet the ellipse in P; and EG and GF together, are greater, and EH and HF together, less, than EP and PF [21. 1 Eu.], or, which is equal [1. 1 Sup], AB. Part 2, fig. 2. Let GF meet the hyperbola, BP in P, and because EG is less than EP and PG together [20. 1 Eu.], the excess of EG above GF, is less than the excess of EP and PG together above GF [Ax. 5. 1 Eu.], or, of EP above PF, or [1. 1 Sup.], AB. And the difference of EH and HF is greater than AB.

For, having drawn HE meeting the hyperbola BP in K, and joined KF; because HF is less than HK and KF together [20. 1 Eu.], the excess of EH above HF, is greater than the excess of EH above HK and KF together, or of EK above KF, or [1. I Sup.], AB.

Part 3, fig. 3. Join GF and HF, and draw GL and HD perpendicu

Figl

A

Fig.2.

E

K

A

B

Fig.31

F

0

H

H

lar to DO; and, because G is without the parabola and F within it, GF meets the parabola, as in P, and, for the same reason, HD meets it, as in K; draw PO perpendicular to DO, and join KF and GO; and PF and PO being equal [Def. 8. 1 Sup.], GF is equal to GP and PO together [Ax. 2. 1 Eu], and therefore, GP and PO together being greater than GO [20. 1 Eu.], and GO than GL[Cor. 1. 19. 1 Eu.], GF is greater than GL. And HF is less than HK, KF together [20. 1 Eu.], or, KB being equal to KF [Def. 8. 1 Sup.], than HD.

PROP. IV. THEOR.

If the sum of the distances of any point from the focuses (E and F, see fig. 1 above), of an ellipse, be equal to the principal axis (AB), that point is in the ellipse; if that sum be greater, it is without, if less, within the same.

If the difference of the distances of any point from the focuses (E and F, see fig. 2 above), of a hyperbola, be equal to the principal axis, that point is in one of the opposite hyperbolas; if that difference be less, it is without both, if greater, within one of them.

And if the distance of any point from the focus (F, see fig. 3 above), of a parabola, be equal to its distance from the directrix (DO), that point is in the parabola; if the distance from the focus be greater, the point is without, if less, within it.

Part 1. Fig. 1. If the sum of EP and PF be equal to AB, P is in the ellipse; for, if P were without the ellipse, that sum would be greater, if within it, less, than AB [3. 1 Sup.], contrary to the supposition.

If the sum of EG and GF be greater than AB, G is without the ellipse; for, if G were in the ellipse, that sum would be equal to, if within it, less than AB [1 and 3. 1 sup], contra hyp.

And if the sum of EH and HF be less than AB, H is within the ellipse; for, if H were in the ellipse, that sum would be equal to, if without it, greater than AB (1 and 3. 1 Sup.), contra hyp.

Part 2, fig. 2. If the difference of EP and PF be equal to AB, P is one of the opposite hyperbolas; for, if P were without both of them, that difference would be less, if within one of them, greater, than AB (3, 1 Sup.), contra hyp.

If the difference of EG and GF be less than AB, G is without both of the opposite hyperbolas; for, if G were in

one of these hyperbolas, that difference would be equal to, if within one of them, greater than, AB (1 and 3. 1 Sup.), contra hyp.

And if the difference of EH and HF be greater than AB, H is within one of the opposite hyperbolas; for, if H were in one of them, that difference would be equal to, if without both of them, less than, AB (1 and 3. 1 Sup.), contra hyp.

Part 3, fig. 3. If PF drawn to the focus be equal to PO drawn perpendicularly to the directrix, P is in the parabola; for, if P were without the parabola, PF would be greater, if within it, less than PO (3. 1 Sup.), contra hyp.

If GF drawn to the focus be greater than GL drawn perpendicularly to the directrix, G is without the parabola; for, if G were in the parabola, GF would be equal to, if within it, less than, GL (1 and 3. 1 Sup.), contra hyp.

And if HF drawn to the focus be less than HD drawn perpendicularly to the directrix, H is within the parabola; for, if H were in the parabola, HF would be equal to,, if without it, greater than, HD (1 and 3. 1 Sup.) contra hyp.

PROP. V. THEOR.

Any diameter of an ellipse or hyperbola is bisected in the centre.

If the diameter be an axis, the Fig1 proposition is demonstrated in 1. 1 Sup.

But if it be any other diameter, A as QP, (see fig. 1 and 2), C being the centre, QP is bisected in C.

For if QC and CP be not equal, let one of them, as CP, if possible, be greater than the other QC, and take on CP, a part CR equal to CQ, and to the focuses E, F, draw PE, PF, QE, QF, RE and RF, and join EF, producing it, in fig. 1, both -ways, to the principal vertices A and B.

The triangles ECR, QCF (see both fig.), having the sides EC, CR and angle ECR, severally equal to FC, CQ and the angle FCQ, the right

R

E

C

B

Fig. 2.

E

H

T

lines ER and FQ are equal [4. 1 Eu.]; for the same reason, FR and EQ are equal, therefore the sum in fig. 1, and difference in fig. 2, of ER and RF is equal to the sum or difference, as the case may be, of FQ and QE, or, which is equal [1. 1 Sup.] of EP, PF. But this is absurd, in fig. 1, ER, RF being less than EP, PF, [21. 1 Eu.], therefore CP is not greater than CQ in that case; in like manner it may be proved, that CP is not less than CQ, therefore CP and CQ are equal, and QP is bisected in C, in the case of the ellipse, fig. 1.

And in the case of the hyperbola, fig. 2, besides the above construction, on PE take PG equal to PF, and on RE, RH equal to RF; join RG and GH, divide EF in K, so that EK may be to KF, as EP to PF [Cor. 1. 10. 6 Eu.], and join PK : and, since the triangles ECP and FCP have the sides EC and CP severally equal to FC and CP, but the obtuse angle ECP is greater than the acute angle FCP, the base EP is greater than the base FP [24. 1 Eu.], therefore EK is greater than KF [constr. and cor. 13, 5 Eu.], therefore EC and CF being equal [Def. 5. 1 Sup.], the point K falls between C and F; and, because EP is to PF, as EK to KF [constr.], the angle EPK is equal to KPF [3. 6 Eu.], therefore the angle EPC is less than CPF; whence, the triangles GPR and FPR having the sides GP and PR, severally equal to FP and PR, but the angle GPR less than FPR, the base GR is less than the base RF [24. 1 Eu.], or its equal RH; therefore the angle RHG is less than RGH [18. 1 Eu.], and therefore acute [32. 1 Eu.], and so the angle EHG is obtuse [13. 1 Eu.], and therefore EGH acute [32. 1 Eu.]; therefore EG, the excess of EP above PF, is greater than EH, the excess of ER above RF [19. 1 Eu.]; but these excesses are above proved to be equal, which is absurd therefore C is not greater than CQ; in like manner it may be proved, that it is not less, therefore CP and CQ are equal, and QP is bisected in C, in the case of the hyperbola, fig. 2.

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