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Make the angle BAD equal to the given angle C; I. Prob. 5. from A draw a straight line perpendicular to AD;

I. Prob. 2. draw a straight line bisecting AB at right angles;

1. Prob. 4. let these two straight lines meet at O; with centre O and radius OA describe the circle ABE: AEB, the segment alternate to BAD, shall be the segment required.

Because the straight line bisecting AB at right angles is the locus of points equidistant from A and B,

Loci, iii. therefore O is equidistant from A and B; therefore B lies on the circumference of the circle whose centre is O and radius OA.

III. 1, Cor. Again, because the angle OAD is a right angle, therefore AD is a tangent to the circle ; therefore the angle BAD is equal to the angle in the alternate segment AEB; but the angle BAD is equal to the angle C;

Constr. therefore the angle in the segment AEB is equal to the given angle C; therefore AEB is the segment required.

Q.E.F.

III. 20.

III. 23.

Ex. 92. Examine the case in which the given angle is a right angle.

PROB. 4. From a given circle to cut off a segment containing a given angle.

Let ABC be the given circle, D the given angle: it is required to cut off from ABC a segment containing an angle equal to the angle D.

B

At any point A on the circumference of the circle draw the tangent AE,

III. 20, Cor. 5. and draw the chord AB, making with AE an angle equal to D:

I. Prob. 5. ACB, the segment alternate to BAE, shall be the segment required.

Because AE is a tangent to the circle, and AB a chord drawn from the point of contact, therefore the angle BAE is equal to the angle in the segment ACB, alternate to BAE

a

III. 23. But the angle BAE is equal to the angle D;

Constr. therefore the angle in the segment ACB is equal to the given angle D.

Q.E.F.

EXERCISES.

а

93. Two equal circles have their centres at A and B, O is a fixed point outside those circles, A is the centre of a third circle whose radius is equal to OB; prove that the tangents from O to the three circles are equal to the sides of a right-angled triangle.

94. Describe a triangle having given the vertical angle, one of the sides containing it, and the altitude.

95. Draw a circle through a given point to touch a given straight line at a given point.

96. Tangents are drawn to the circumscribing circle of an isosceles triangle through the angular points. Shew that they form an isosceles triangle. Shew also that the two triangles cannot have equal vertical angles unless they are both equilateral.

97. If a parallelogram be inscribed in, or circumscribed about a circle, its diagonals pass through the centre.

SECTION VI.

Two CIRCLES.

DEF. 14. Two circles are said to touch externally at a point where they meet, if each lies outside the other ; to intersect if a part of each lies inside, and the remaining part outside the other; and to

' touch internally at a point where they meet, if one of them lies inside the other.

THEOR. 24.

If two circles meet in a point which is not on the line joining their centres, they meet in one other point; and the circles intersect; the line joining the two points of intersection is bisected at right angles by the line joining their centres; and the distance between the centres is greater than the difference, and less than the sum, of the radii.

Let the circumference of two circles whose centres are A and B meet in the point C, not on the straight line through A and B:

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Join AB.

Draw CE perpendicular to AB, and produce CE to D, so that ED may be equal to CE.

1. Prob. 3.

In the triangles AEC, AED, the side CE is equal to the side DE, the side AE is common to both, and the angle AEC is equal to the angle AED; therefore the side AC is equal to the side AD;

1. 5. therefore D is a point on the circumference of the circle whose centre is A.

III. 1, Cor.

In like manner, D is a point on the circumference of the circle whose centre is B. Therefore the circles meet in a second point D, and they cannot meet again.

III. 12, Cor. 2.

Also the circles shall intersect.

Let the line through A and B cut the circle whose centre is A in F and H, and the other circle in G and K, F being on the same side of A as B, and G on the same side of B as A.

Because BK is equal to BC, therefore AK is equal to the sum of AB and BC; therefore AK is greater than AC; therefore K is outside the circle whose centre is A.

I. 13.

III. 1, Cor.

Ax. e.

Again, because BG is equal to BC; therefore AG is equal to the difference of AB and BC. But the difference of AB and BC is less than the third side AC of the triangle ABC,

5

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