PROPOSITION 1. To find the centre of a given circle. Let ABC be the given ; it is required to find its centre. at G E Take any pt. G within the O, but not in CE, and join GA, GD GB. In the two As ADG, BDG AD=BD, GD is common, and ADG is unequal to BDG; ... GA is unequal to GB, ... G is not the centre. [I. 24. In the same way it may be shown that any other pt. within the outside CE is not the centre; ... F, the mid-pt. of CE, must be the centre. COROLLARY.-From this it is clear that If in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. In other words If a circle pass through two given points, its centre lies on the perpendicular bisector of their join. Hence If any number of circles pass through two given points their centres all lie on the perpendicular bisector of the join of the two given points. NOTES. When a straight line is spoken of as 'drawn within a circle,' it is implied that its extremities are on the circumference. Any straight line, such as AB, drawn within a circle, is called a chord of the circle. Hence the Corollary might be enunciated thus : The perpendicular bisector of any chord of a circle passes through the centre of the circle. Ex. 265.-A circle cannot have two centres. Ex. 266.-Circles whose centres are A and B intersect in C. Through C draw a line PCQ parallel to AB, terminated by the circles. Show that PQ=2 AB. Show that The perpendicular bisectors of CP, CQ pass through A and B respectively. Ex. 267.-ABCD is a quadrilateral inscribed in a circle. the perpendicular bisectors of its sides and its diagonals are concurrent. Extend the theorem to other rectilineal figures. Ex. 268.-A, B, C, D are four points. Show that if the perpr. bisectors of AB, AC, AD are concurrent, so also are the perpendicular bisectors of BC, BD, CD. Extend the theorem to a system of five points. It will be noticed that the word 'circle' is used in two different senses. Sometimes it denotes the figure bounded by the circumference, as when we speak of the segment of a circle (see III. 21); sometimes the circumference itself, as when we speak of a circle passing through two given points. According to Euclid's definition it should be used in the first sense only, but we are not aware of any writer whose use of the word is thus strictly limited. Euclid himself, for instance, speaks (I. 1) of a point where two circles cut each other, when he means a point where their circumferences cut. No confusion, however, can well happen from this double use of the word, and it is certainly convenient in many cases to avail ourselves of it. There are several other instances in Mathematics of this double use of a word in cases where no mistake is likely to follow from it. The symbol for circumference is 'Oce.' PROPOSITION 2. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let A and B be two pts. on the Oce of a O. Let C be the centre of the O, and P any point in AB. Join СА, СР, СВ. Similarly it may be shown that any other pt. in AB, except its extremities, lies within the O, ... AB lies within the O. Ex. 269.-The extremities of the base of an isosceles triangle are farther from the vertex than any other point in the base. PROPOSITION 3. (1) If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre it shall cut it at right angles. (2) If a straight line drawn through the centre of a circle cut a straight line in it which does not pass through the centre at right angles it shall bisect it. (1) Let ABC be a O, and let a st. line CD, passing through the centre E, bisect a st. line AB which does not pass through the centre, at F, then CD is ir to AB. B Join EA, EB. In the As AFE, BFE EF is common, and rad. EA=rad. EB, i.e. CD is Lr to AB. (2) Next let CD, passing through the centre E, cut AB at rt. s at F, then AF=FB. With the same construction, rad. EA rad. EB, .. LEAB= = LEBA. |