Let DQ be the position of DP in this case.
Draw AE perpendicular to DQ.

Because AE is equal to the radius of the circle, therefore E is a point on the circumference.

But QED is drawn through E at right angles to AE, therefore QED touches the circle,

and it passes through D by hypothesis.

By similar reasoning another line DF may be drawn through D touching the circle in the point F on the other side of AD. Then in the two right-angled triangles DEA and DFA we have

AE equal to AF and the hypothenuse AD common, therefore the triangles are equal in all respects, therefore DE is equal to DF,

and ADE is equal to ADF.

PROPOSITION 21.

If a straight line touch a circle and from the point of contact another straight line be drawn cutting the circle, the angles which the cutting line makes with the opposite parts of the touching line shall be equal to the angles in the alternate segments of the circle.

Fig. 31.

B

C

Through the point A of the circle ABC let the straight line EF be drawn touching the circle, and from A let AC be drawn cutting the circle, then shall the angles EAC and FAC be equal to the angles in the segments CDA and CBA respectively.

From A draw AB at right angles to EF meeting the circle again in B, join BC, take any point D in the segment CDA, and join DA, DC, and DB.

E

F

D

Because EF touches the circle at the point A, and AB

is at right angles to EF,

therefore AB coincides with the diameter of the circle passing through the point A (Bk. II. Prop. 14, Cor.),

therefore BCA is a semicircle.

Because BCA is a semicircle therefore the angle ACB is

a right angle.

Because ACB is a right angle therefore CBA+BAC is equal to a right angle,

therefore CBA+BAC is equal to BAF,

therefore CBA is equal to BAF~BAC,

that is, CBA is equal to CAF,

or CAF is equal to any angle in the segment CBA. Because CBA+CDA is equal to two right angles (Bk. II. Prop. 12),

and also CAF+CAE is equal to two right angles (Bk. I. Prop. 11),

therefore CBA+CDA is equal to CAF+CAE,
but CBA is equal to CAF,
therefore CDA is equal to CAE.

EXAMPLES.

1. Two circles cut one another in the points A and B. AC and AD are drawn, each touching one circle, and terminated by the other circle in the points C and D. Prove that the triangles ABC and ABD are equi-angular to one another.

2. On the same base and on the same side of it there are two segments of circles, of which ACB is a semicircle and ADB a quadrant. Through P, any point in ADB, the line APQ is drawn, cutting ACB in Q, and QB is joined. Prove that PQ is equal to QB.

3. ACB and ADB are two segments of circles on the same base AB; AC and BC are two lines from A and B meeting at C on the segment ACB, and produced to meet the segment ADB in D and E respectively. Prove that wherever C may be the arc DE is of constant length.

4. If two circles intersect, prove that the common chord, when produced, bisects each of the common tangents.

5. If two circles touch each other externally, and parallel diameters be drawn one to each circle, prove that the straight line which joins the extremities of these diameters towards opposite parts passes through the point of contact.

6. ABC is a triangle inscribed in a circle O whose centre is joined to the middle point D of the arc BC, and AD is drawn. Prove that the angle ADO is half the difference of the angles B and C.

7. Two circles intersect in A, and through A the secants ABC and ADE are drawn. Prove that the angle between the straight lines BD and CE is constant.

8. Two fixed lines AB and AC are cut by a movable line BC in such a way that the perimeter of the triangle ABC is constant. Prove that BC always touches a certain circle.

9. Two circles intersect in A and B, and CBD is drawn perpendicular to AB, to meet the circles in C and D. If EAF bisect the exterior angle between CA and DA prove that the tangents to the circles at E and F intersect on a point in AB produced.

10. Given a circle O and an exterior point A. A circle is described with A as centre and OA as radius, and another with O as centre, but with double the radius of the circle O. The latter cuts the former in B and C, and the lines OB and OC cut the original circles in E and D. Prove that AE and AD touch this circle.

11. An acute-angled triangle is inscribed in a circle, and the paper on which it is drawn is folded along each of the sides of the triangle. Show that the circumference of the three segments will pass through the same point.

SECTION IV.-ON LOCI CONNECTED WITH THE CIRCLE.

PROPOSITION 22.

The locus of a point at which a given straight line subtends a given angle is two circular arcs, each passing through the extremities of the given line.

Let BC be the given straight line and P a point on the required locus, and therefore such that the angle BPC is equal to the given angle.

B

P

Fig. 32.

Through the points B, P, and C, one, and only one, circumference can be drawn (Bk. II. Prop. 5, Cor.); let this be BPC.

Because the angles in the same segment of a circle are equal to one another, if any point Q, different the circumference BPC, the angle

from P, be taken in
BQC will be equal to the angle BPC,

therefore every point on the arc BPC is a point of the required locus.

Again, let any other point be taken, as R, without the arc BPC and let S be the point in which the line BR meets this

arc.

Then because the exterior angle of a triangle is greater than the interior non-adjacent angle, therefore BSC is greater than BRC.

Therefore BRC is less than the given angle.

Therefore no point above BC and without the arc BPC can be on the required locus.

In like manner, it may be shown that no point above BC and within the arc BPC can be on the required locus.

Therefore no point above BC, and not situated on the arc BPC, can be on the required locus.

An arc equal to BPC can be described on BC and below BC.

Therefore the locus in question is two circular arcs, each passing through B and C.

Corollary. When the given angle is a right angle the locus is the circle of which BC is a diameter.

PROPOSITION 23.

The locus of the middle points of equal chords in a given circle is another concentric circle.

Let O be the centre of the circle.

Let AB be one of the chords of given length, and P its

middle point.

Let CD be any other such chord, and Q its middle point.

Join OP and OQ, then these lines are perpendicular to AB and CD respectively (Bk. II. Prop. 3, Cor. 3).

But equal chords in a circle are at equal distances from the centre,

therefore OP is equal to OQ. Therefore the middle point of

every chord equal to AB is on the circumference of a circle passing through P, and having its centre at O, and conversely because lines equally distant from the centre are equal, it follows that all points on this circle are middle points of equal chords.

PROPOSITION 24.

The locus of the middle points of all chords drawn through a given point in the circumference of a given circle is a circle passing through the given point and having the radius of the given circle to the given point for a diameter.

Let APB be the given circle, C the centre, and A the given point on the circumference.