Hence the angles MPQ, PSQ, being equal to supplements of equal angles, are equal (I. 51).

APPLICATIONS OF THEOREM 28

77. PROBLEM 4. On a given line, to describe a segment of a circle containing an angle equal to a given angle.

Let AB be the given line, and C the given angle.

To describe on AB a segment of a circle containing an angle equal to C.

Draw AD, making the angle BAD equal to the given angle c (I. 77).

Draw 40 perpendicular to AD; and draw MO bisecting AB at right angles. Let these lines meet in O; and draw OB.

N

M

The lines OA and OB are equal (I. 64).

b

B

Therefore the circle described with o as center, and with radius 04, passes through B. Let ABN be this circle.

The segment ABN, alternate to the angle BAD, is the required segment.

Since OAD is a right angle, AD touches the circle (63). Therefore the angle BAD is equal to the angle in the alternate segment ABN (76).

Hence the angle in the segment ABN is equal to C.

78. Ex. Consider the case in which the angle C is a right angle.

79. Cor. If the base and vertical angle of a triangle are given, the locus of its vertex consists of the arcs of the two segments described on the base, containing an angle equal to the given vertical angle.

Show that a triangle on the given base satisfies the requirements if its vertex is on one of these arcs, and not otherwise.

80. PROBLEM 5. From a given circle, to cut off a segment containing an angle equal to a given angle. Let ABC be the given circle, and D the given angle.

angle FBC equal to the given angle D (I. 77).

The segment BAC, alternate to the angle FBC, is the segment required.

Since BF is a tangent, the angle FBC is equal to the angle in the alternate segment BAC (76).

Therefore the angle in the segment BAC is equal to the given angle D.

Discussion. How many solutions are there to this problem? How many solutions will there be if the statement of the problem is modified in the following manner: Through a given point on a circle, to draw a line that shall cut off a segment containing an angle equal to a given angle.

EXERCISES

1. All chords of a circle that touch a concentric circle are equal. 2. Through a given point draw a line so that the chord intercepted by a given circle shall be equal to a given line. [Use ex. 1 and art. 74. State the restrictions on the data when the point is within the circle; also when the point is without or on the circle.]

3. The part of any tangent intercepted by two parallel tangents subtends a right angle at the center of the circle.

4. If through the center of a circle two perpendicular lines are drawn to meet any tangent, then the tangents drawn from the two points of intersection are parallel.

5. To draw a tangent to a given circle making a given angle with a given line.

6. Any chord of a circle bisects the angle between the diameter through one extremity and the perpendicular from it on the tangent at the other.

7. Draw a circle through a given point to touch a given line at a given point. [Use 3 and 67.]

8. If a quadrangle is circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair.

9. If a convex quadrangle is such that the sum of one pair of opposite sides is equal to the sum of the other pair, then a circle may be inscribed in it.

81. Definitions.

TWO CIRCLES

Two circles are said to intersect at a point where they meet if they cross each other at this common point.

Two circles are said to touch at a point where they meet if they do not cross each other at this common point; and this point is called the point of contact.

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82. The line passing through the centers of two circles is called their central line.

POINTS COMMON TO TWO CIRCLES

Common point not on central line.

83. THEOREM 29. If two circles have one common point, not on their central line, then they have a second common point; and the circles intersect at each of these two points.

Let two circles whose centers are O and o' have the common point P, not on their central line.

R

First, to prove that they have a second common point. Draw PN perpendicular to 00', and prolong it to Q making NQ equal to PN. Draw OP, 0Q, O'P, O'Q.

By equality of triangles it follows that OP equals OQ, and O'P equals O'Q.

Therefore Q is a point on each of the circles (6).

Moreover, there is no third common point (22).

Next, to prove that the circles intersect at each of the points P, Q.

Let R, S be any two points on the circle whose center is o', situated at opposite sides of the point P. Draw OR, O'R, OS, O's.

In the triangles OO'R and 00'P, the sides oo' and O'R are respectively equal to the sides oo' and O'P, and the included angle 00'R is greater than 00'P.

Therefore the third side OR is greater than the third side OP (I. 91).

Therefore the point R is without the circle whose center is 0 (6).

In a similar way it is proved that os is less than OP. Therefore the point s is within the circle whose center is 0.

Now R and S are any two points on the circle whose center is o', situated at opposite sides of P.

Hence the two circles cross each other at P.

Similarly it can be proved that they cross at Q.

Common point on central line.

84. THEOREM 30. If two circles have one common point, situated on their central line, then they have no other common point; and the circles touch at this point.

Let 0, o' be the centers of the two circles, and P the common point on the central line 00'.

First, to prove that there is

no other common point on the central line.

Suppose, if possible, that

is

another point on the line 00', common to the two circles.

R

Then the segment PQ is a diameter of each circle. Hence the middle point of PQ is the center of each circle; which is impossible since the centers o, o' do not coincide.

Therefore there is no second common point on the central

line.

Next, to prove that there is no second common point not on the central line.

Suppose, if possible, that R is another common point not on the central line. Then there is a third common point R', not on the central line (83).

Since there are three common points, the two circles coincide throughout (21).

This is contrary to the hypothesis; therefore there is no common point not on the central line.

Hence P is the only common point.

Again, to prove that the circles touch at P.

Let R be any other point on the circle whose center is o'. Draw OR and O'R.

Since O'R equals O'P, therefore the sum of O'R and 00' is equal to OP.

But OR is less than the sum of 00' and O'R (I. 87).