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SIMSON'S LINE. 74. If from any point on the circumference of the circle circumscribed about a triangle, perpendiculars are drawn to the three sides, the feet of these perpendiculars are collinear.
Let P be any point on the Oce of the circle circumscribed about the A ABC; and let PD, PE, PF be the perps drawn from P to the three sides.
It is required to prove that the points D, E, F are collinear.
Join FD and DE:
Join PB, PC.
:. the points P, D, B, F are concyclic:
III. 21. But since BACP is a quad' inscribed in a circle, having one of its sides AB produced to F,
.. the ext. Ź PBF =the opp. int. L ACP. Ex. 3, p. 202.
To each add the L PDE:
:: the points P, D, E, C are concyclic;
:: FD and DE are in the same st. line;
that is, the points D, E, F are collinear. Q.E.D. [The line FDE is called the Pedal or Simson's Line of the triangle ABC for the point P; though the tradition attributing the theorem to Robert Simson has been recently shaken by the researches of Dr. J. S. Mackay.]
75. ABC is a triangle inscribed in a circle ; and from any point P on the circumference PD, PF are drawn perpendicular to BC and AB: if FD, or FD produced, cuts AC at Ė, shew that PE is perpendicular to AC.
76. Find the locus of a point which moves so that if perpendiculars are drawn from it to the sides of a given triangle, their feet are collinear.
77. ABC and AB'C' are two triangles having a common vertical angle, and the circles circumscribed about them meet again at P; shew that the feet of perpendiculars drawn from P to the four lines AB, AC, BC, B'C' are collinear.
78. A triangle is inscribed in a circle, and any point P on the circumference is joined to the orthocentre of the triangle : shew that this joining line is bisected by the pedal of the point P.
ON THE CIRCLE IN CONNECTION WITH RECTANGLES.
[See Propositions 35, 36, 37.]
1. If from any external point P two tangents are drawn to a given circle whose centre is O, and if OP meets the chord of contact at Q; then the rectangle OP, OQ, is equal to the square on the radius.
Let PH, PK be tangents, drawn from the external point P to the O HAK, whose centre is O; and let OP meet HK the chord of contact at Q, and the Oce at A. Then shall the rect. OP, OQ=the sq. on OA.
On HP as diameter describe a circle : this circle must pass through Q, since the & HQP is a rt. angle.
III. 16. :: the rect. OP, OQ=the sq. on OH,
III. 36. =the sq. on OA. Q.E.D.
2. ABC is a triangle, and AD, BE, CF the perpendiculars drawn from the vertices to the opposite sides, meeting in the orthocentre O: shew that the rect. AO, OD=the rect. BO, DE=the rect. CO, OF.
3. ABC is a triangle, and AD, BE the perpendiculars drawn from A and B on the opposite sides: shew that the rectangle CA, CE is equal to the rectangle CB, CD.
4. ABC is a triangle right-angled at C, and from D, any point in the hypotenuse AB, a straight line DE is drawn perpendicular to AB and meeting BC at E: shew that the square on DĖ is equal to the difference of the rectangles AD, DB and CE, EB.
5. From an external point P two tangents are drawn to a given circle whose centre is O, and OP meets the chord of contact at Q: shew that any circle which passes through the points P, Q will cut the given circle orthogonally. [See Def. p. 240.]
6. A series of circles pass through two given points, and from a fixed point in the common chord produced tangents are drawn to all the circles : shew that the points of contact lie on a circle which cuts all the given circles orthogonally.
7. All circles which pass through a fixed point, and cut a given circle orthogonally, pass also through a second fixed point.
8. Find the locus of the centres of all circles which pass through a given point and cut a given circle orthogonally.
9. Describe a circle to pass through two given points and cut a given circle orthogonally.
10. A, B, C, D are four points taken in order on a given straight line : find a point between B and C such that the rectangle OA, OB may be equal to the rectangle OC, OD.
11. AB is a fixed diameter of a circle, and CD a fixed straight line of indefinite length cutting AB or AB produced at right angles ; any straight line is drawn through A to cut CD at P and the circle at Q: shew that the rectangle AP, AQ is constant.
12. AB is a fixed diameter of a circle, and CD a fixed chord at right angles to AB; any straight line is drawn through A to cut CD at P and the circle at Q : shew that the rectangle AP, AQ is equal to the square on AC.
13. A is a fixed point, and CD a fixed straight line of indefinite length; AP is any straight line drawn through Å to meet CD at P; and in AP a point Q is taken such that the rectangle AP, AQ is constant : find the locus of Q.
14. Two circles intersect orthogonally, and tangents are drawn from any point on the circumference of one to touch the other : prove that the first circle passes through the middle point of the chord of contact of the tangents. [Ex. , p. 251.]
15. A semicircle is described on AB as diameter, and any two chords AC, BD are drawn intersecting at P: shew that
AB2=AC. AP + BD.BP. 16. Two circles intersect at B and C, and the two direct common tangents AE and DF are drawn: if the common chord is produced to meet the tangents at G and H, shew that GH2=AE2+ BC?.
17. If from a point P, without a circle, PM is drawn perpendicular to a diameter AB, and also a secant PCD, shew that
PM2=PC. PD+AM. MB.
18. Three circles intersect at D, and their other points of intersection are A, B, C; AD cuts the circle BDC at E, and EB, EC cut the circles ADB, ADC respectively at F and G: show that the points F, A, G are collinear, and F, B, C, G concyclic.
19. A semicircle is described on a given diameter BC, and from B and C any two chords BE, CF are drawn intersecting within the semicircle at O; BF and CE are produced to meet at
; shew that the sum of the squares on AB, AC is equal to twice the square on the tangent from A together with the square on BC.
20. X and Y are two fixed points in the diameter of a circle equidistant from the centre C: through X any chord PXQ is drawn, and its extremities are joined to Y; shew that the sum of the squares on the sides of the triangle PYQ is constant. [See p. 161, Ex. 24.]
PROBLEMS ON TANGENCY.
21. To describe a circle to pass through two given points and to touch a given straight line.
Let A and B be the given points, and CD the given st. line.
It is required to describe a circle to pass through A and B and to touch CD.
Join BA, and produce it to meet
D rect. PA, PB;
II. 14. and from PD (or PC) cut off PQ equal to a side of this square.
Through A, B, and Q describe a circle. Ex. 4, p. 171. Then since the rect. PA, PB=the sq. on PQ, :: the O ABQ touches CD at Q.
III. 37. Q.E.F.
NOTES. (i) Since PQ may be taken on either side of P, it is clear that there are in general two solutions of the problem.
(ii) When AB is parallel to the given line CD, the above method is not applicable. In this case a simple construction follows from III. 1, Cor. and III. 16, and it will be found that only one solution exists.
22. To describe a circle to pass through two given points and to touch a given circle. Let A and B be the given
R points, and CRP the given circle. It is required to describe a circle to pass through A and B, and to touch the O CRP.
Through A and B describe any circle to cut the given circle at P and Q.
Join AB, PQ, and produce them to meet at D.
From D draw DC to touch the given circle, and let C be the point of contact.
Then the circle described through A, B, C will touch the given circle.
For, from the O ABQP, the rect. DA, DB=the rect. DP, DQ : and from the OPQC, the rect. DP, DQ=the sq. on DC;
III. 36. :: the rect. DA, DB=the sq. on DC: :: DC touches the O ABC at C.
III. 37. But DC touches the O PQC at C;
Constr. :. the O ABC touches the given circle, and it passes through the given points A and B.
Q.E.F. NOTE. (i) Since two tangents may be drawn from D to the given circle, it follows that there will be two solutions of the problem.
(ii) The general construction fails when the straight line bisecting AB at right angles passes through the centre of the given circle : the problem
then becomes symmetrical, and the solution is obvious.
23. To describe a circle to pass through a given point and to touch two given straight lines.
Let P be the given point, and AB, AC the given straight lines. It is required to describe a circle
R to pass through P and to touch
S AB, AC.
É Now the centre of every circle which touches AB and AČ must lie on the bisector of the L BAC.
Ex. 7, p. 197.
From P draw PK perp. to AE, and produce it to P', making KP' equal to PK.