FEP, being For, through the point of contact E draw FE, and ƒE meeting P produced in G. Then, the GEP = being each equal to the feh, and the angles at right, and the side PE being common, the two triangles GEP, FEP are equal in all respects, and so GE = FE, and GP FP. Therefore, since FPFG, and Fc = Ff, and the angle at F common, the side CP will be = fG or AB, that is CP = СА ОГ СВ. And in the same manner ch = CA or CB. Q. E. D. Corol 1. A circle described on the transverse axis, as a diameter, will pass through the points P, p; because all the lines CA, CP, cp, CB, being equal, will be radii of the circle. Corol. 2. CP is parallel to fe, and ch parallel to FE. Corol. 3. If at the intersections of any tangent, with the circumscribed circle, perpendiculars to the tangent be drawn, they will meet the transverse axis in the two foci. That is, the perpendiculars PF, f give the foci F, f. THEOREM XIV. The equal Ordinates, or the Ordinates at equal Distances from the Centre, on the opposite Sides and Ends of an Ellipse, have their Extremities connected by one Right Line passing through the Centre, and that Line is bisected by the Centre. That is, if CDCG, or the ordinate DE = GH; For when CD = CG, then also is DE GH by cor. 2, th. 1. But the LD LG, being both right angles; = therefore the third side cE = CH, and the DCE = GCH, and consequently ECH is a right line. Carol. Corol. 1. And, conversely, if ECH be a right line passing through the centre; then shall it be bisected by the centre, or have CE = CH; also DE will be = GH, and CD = CG. Corol. 2. Hence also, if two tangents be drawn to the two ends, E, H of any diameter EH; they will be parallel to each other, and will cut the axis at equal angles, and at equal dis tances from the centre. For, the two CD, CA being equal to the two CG, CB, the third proportionals CT, cs will be equal also; then the two sides CE, CT being equal to the two CH, cs, and the included angle ECT equal to the included angle HCs, all the other corresponding parts are equal: and so the Ts, and TE parallel to HS. Corol 3. And hence the four tangents, at the four extremities of any two conjugate diameters form a parallelogram circumscribing the ellipse, and the pairs of opposite sides are each equal to the corresponding parallel conjugate diameters. For, if the diameter eh be drawn parallel to the tangent TE or нs, it will be the conjugate to EH by the definition; and the tangents to e, h will be parallel to each other, and to the diam eter EH for the same reason. THEOREM XV. If two Ordinates ED, ed be drawn from the Extremities E, e of two Conjugate Diameters, and Tangents be drawn to the same Extremities, and meeting the Axis produced in T and R ; Then shall CD be a mean proportional between cd, dr, and cd a mean proportional between CD, DT. Hence co: cd :: CR: CT. Hence also CD: cd: de DE. cd. de, or A CDE = A cde. Corol. 1. Corol. 3. Also cd2 = CD. DT, and CD2 = cd. dr.. Or cd a mean proportional between CD, DT; THEOREM XVI. The same Figure being constructed as in the last Theorem each Ordinate will divide the Axis, and the Semi-axis added to the external Part, in the same Ratio. Corol. 1. Hence, and from cor. 3 to the last, it is, cd2 = CD. DT = AD. DB = CA2 -CD2, CD2 cd. dR = Ad. dв = CA2 - cd. Corol. 2. Hence also, CA2 = CD2 + cd2, and ca2 DE2 + de2. Q. E. D. Corol. 3. Further, because CA2: ca2:: AD. DB or cd2: DE2, therefore CA: ca: cd: DE. likewise CA ca: CD: de. THEOREM XVII. 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: Will be to the Dist. PD :: As Sq. of the Trans. Axis: DP C CA2: ca2:: DC: DP. Fer, For, by theor. 2, CA2 : ca2 :: AD. DB: DE, But, by rt angled As, the rect. TD . DP = DE2; therefore or AC: ca: DC : DP. Q. E. D. THEOREM XVIII. 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. CE: CN; CA: CT; By sim. triangles, CD: CA: but, by theor. 7, CD CA: theref. by equal. CACT:CE: CN. 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 = Corol. 1. From each of the equal tri. CET, CAN, take the common space CAPE, A CAN. and there remains the external ▲ PAT = ▲ PNE. Corol. 2. Also from the equal triangles CET, CAN, take the common triangle ĈED, and there remains the ▲ TED = trapez. ANED. THEOREM XIX. 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 AKQG trapez. ANHG, and A Kg-trapez. ANhg. K AS D B For draw the ordinate De. Then The three sim. triangles CAN. CDE, CGH, are to each other as .CA2, CD2, CG; th. by div. the trap.ANED: trap. ANHG:: CA-CD2 CA2-CG. But, by theor. 1, DE2: GQ:: CA2-CD2: CA-eG", theref. by equ. trap. ANED: trap. ANHG: DE2 but, by sim As, tri TED: tri. theref, by equality, ANED: KQG DE TED ::ANHG :GQ. : KQG. But, by cor. 2, theor. 18, the trap. ANED A TED; Corol. Corol. 2. From the equals ANHG, KQG, take the equals ANhg, Kqg, i and there remains ghHGgqQG. Corol. 3. And from the equals ghнG, gyQG, take the common space gqLHG. and there remains the ▲ LQH = ▲ Lgh. Corol. 4. Again from the equals KQG. TEHG, TELKA LQH. R the triangle KQG becomes the triangle IRC, Q. E. D. M B Corol. 6. Also when the lines KQ and HQ, by moving with a parallel motion, come into the position ce, Me, the |