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Therefore if R be the point of intersection of the segments, OR is perpendicular to' MV, and since MV bisects OC, therefore MN is parallel to RC.
Therefore the angle ORC is a right angle, and therefore the segments intersect on a circle described on OC as diameter.
iv. Describe an isosceles triangle having each of the angles at the base double of the third angle.
Prove that the circle drawn through the middle points of the sides of this triangle will intercept portions of the equal sides such that a regular pentagon can be inscribed in the circle having these portions as two of its sides.
If D, E, F (fig. 4) be the middle points of the sides, the triangle FBD will be similar to the triangle ABC, and he circle through D, E, F will be the small circle of Euclid's construction relatively to the triangle FBD, and will cut AB, AC in points MN, such that
BD=DN=NF= FE, and therefore DMEFN is a regular pentagon.
v. Equal triangles which have one angle of the one equal to one angle of the other have their sides about the equal angles reciprocally proportional; and triangles wbich have one angle of the one equal to one angle of the other and their sides about the equal angles reciprocally proportional are equal to one another
If ABC, ADE be two such triangles placed so that BA, AE are in a straight line, as also CA and AD; and if BC, DE produced meet in F, prove that FA will bisect CE and BD.
Since (fig. 5) the BAC =- DAE; therefore
OBCE=A DCE, and therefore CE is parallel to BD; therefore
BC: BF :: DE: DF; therefore A BCA : - BFA :: ADEA:' DFA,
ABCANA DEA; therefore
ABFA =- DFA, therefore perpendiculars from B and D on FA are equal, and therefore FA bisects BD, and its parallel CE.
vi. If a straight line stand at right angles to each of two straight lines at the point of their intersection, it shall also be at right angles to the plane which passes through them, that is to the plane in which they are.
If P be a point equidistant from the angles A, B, C of a right-angled triangle of which A is the right angle and D the middle point of BC, prove that PD is at right angles to the plane of ABC. Prove also that the angle between the planes PAC, PBC and the angle between the planes PAB, PBC are together equal to the angle between the planes PAC, PAB.
For (fig. 6) PA = PBP = PD? + DBP = PD? + DA”; therefore PDA is a right angle, and since PDB is also a right angle, therefore PD is at right angles to the plane ABC.
And since PA = PB and DA= DB, therefore the angle between the planes PAB, PBC is equal to the angle between the planes PAD, PAB.
Similarly the angle between the planes PAC, PBC is equal to the angle between the planes PAD, PAC.
Therefore the angle between the planes PAB, PBC, and the angle between the planes PAC, PBC are together equal to the angle between the planes PAC, PAB.
Compare the rider to question 2, Thursday afternoon, Jan. 7.
vii. The ordinate to the diameter through any point on a parabola is a mean proportional between the corresponding abscissa and four times the focal distance of that point.
If through a fixed point A a straight line be drawn meeting two fixed lines OD, OE in B and C respectively, and on it a point P be taken such that AC.AP= AB”; prove that the locus of P is a parabola which passes through
A and 0 and has its axis parallel to OD and the tangent at A parallel to OE.
Let the straight line through A (fig. 7) parallel to OD meet OE in F, and draw PM parallel to OE to meet AF in M.
By similar triangles
PM? : OF? :: AP? : AB? :: AP: AC:: AM: AF; therefore the locus of P is the parabola stated.
viii. If the tangent and ordinate at any point P of an ellipse meet the axis major in T and N respectively, then CT.CN= CA.
If any circle be drawn through N and T, prove that it is cut at right angles by the auxiliary circle of the ellipse. Let the circle cut the circumscribing circle in P, then
CT.CN= CA’ = CP%; therefore CP is a tangent to the circle, and therefore it cuts the auxiliary circle at right angles.
ix. Tangents to an ellipse or hyperbola at right angles to each other intersect on a fixed circle.
If any rectangle circumscribe an ellipse, prove that the perimeter of the parallelogram formed by joining the points of contact is equal to twice the diameter of the circle which is the locus of the point of intersection of tangents at right angles.
If PQRS (fig. 8) be a circumscribing rectangle and C the centre of the ellipse.
Join CP and let it cut LM one side of the parallelogram in N.
C is the centre of the rectangle and of the parallelogram and P, Q lie on the director circle; therefore A CPQ is isosceles, hence so also is » LNP; therefore LN=NP, and CN is parallel to another side of parallelogram and equal to half of it; therefore CP= 4 perimeter of parallelogram or perimeter of parallelogram = 4CP= twice diameter of director circle.
10. In a central conic the tangent makes equal angles with the focal distances, and the sum or difference of the focal distances is constant.
Given a focus, the length of the transverse axis and that the second focus lies on a fixed straight line, prove that the conic will touch two fixed parabolas having the given focus for focus.
If S (figs. 9 and 10) be the given point, CH the given straight line, draw DK, D'K' parallel to CH at distances equal to the given transverse axis, and draw SC perpendicular to CH to meet it in C, and DK, D'K' in D and D'.
Then if H be the second focus of a conic and PP' be drawn throngh H parallel to SC to meet the conic in P, P' and DK, D'K' in K, K', it is evident that
SP= PK and SP = P'K'. Therefore the conic touches two fixed parabolas having the common focus S and the directrices DK, D'K' at P and P.
If SC (fig. 9) is less than the given transverse axis, the parabolas are turned in opposite directions and intersect at right angles on CA in B and B', such that SB= SB' = given transverse axis.
Then if H be taken between B and B', the conic is an ellipse; but if H be taken beyond Bor B', the conic is a hyperbola, and the same branch of it touches the parabolas.
'If sc' (fig. 10) is greater than the given transverse axis, the parabolas are turned in the same direction and do not intersect, and the conic is always a hyperbola of which different branches touch the parabolas.
11. The tangents to a conic section drawn from a point subtend equal angles at either focus.
If PP' be a chord of a conic parallel to the transverse axis and the two circles be drawn through a focus S touching the conic at P and P' respectively, prove that F the second point of intersection of the circles will be at the intersection of PP and ST, where T is the point of intersection of the tangents at P and P'.
Prove also that the locus of F for different positions of PP' will be a parabola with its vertex at S.
A circle can be described about TPSP' (fig. 11) and therefore 2 TPP' =_ FSP; therefore the circle described on SP touching PT at P passes through F.
Similarly the circle described on SP touching P'T at P' passes through F.
Also if FM, FN be drawn perpendicular to the transverse and conjugate axes to meet them in M, N respectively, CT.CN=CB”; therefore
FW: CB2 :: CN: CT:: SM: SC; therefore the locus of F is a parabola with its vertex at S and passing through the ends of the conjugate axis.
12. If a right circular cone be cut by a plane, the distance of any point on the curse of section from a certain point bears a constant ratio to the distance from a certain straight line.
If any sphere be inscribed in the cone, the length of the tangent line drawn from any point on the curve of section to the sphere will bear the same constant ratio to the distance of the point from the line of intersection of the plane section and the plane of the circle of contact of the sphere and cone.
The rider is proved incidentally in the book work.
(If O be the centre of the sphere and OD be drawn perpendicular to the plane of the section, it can easily be shewn that the length of any tangent line from a point in the plane to the sphere is equal to the distance of this point from a fixed point S in OD such that SD = OD" - (rad. of sphere)”.
The locus of S is called the focal conic of the conic section.)