70. Two of the straight lines which join the extremities of equal arcs are parallel to one another.

Because arc A C = arc B D,

LBAD LCDA.

Since these are alternate

angles, therefore A B and CD

are parallel.

E

SECTION II.-Angles in a Circle.

71. The straight lines which join the opposite extremities of two parallel chords intersect in that diameter which is perpendicular to the chords.

Let A B, C D be the parallel chords, and let the straight lines A D, C B intersect in the point P (see preceding fig.). Through P draw E F perpendicular to A B, then we have to prove that EF is a diameter.

Because all angles at the circumference on the arc BD are equal,

.. ▲ BAD = ▲ B C D,

But because A B and C D are parallel,

Hence ▲ APB is isosceles and the perpendicular from the apex to the base bisects the base; therefore E F bisects the chord A B at right angles and is a diameter.

72. If two chords of a circle intersect within the circle, the angle between them is equal to half the angle at the centre on the sum of the intercepted

arcs.

Let A B, C D be chords intersecting in E. Join

equal to ▲ A+ D is equal to half the angle at the centre on the sum of the arcs A C, B D.

73. If two chords of a circle intersect without the circle, the angle between them is equal to half the angle at the centre on the difference of the intercepted

arcs.

Let A B and C D be chords intersecting in E outside the circle. Join B D.

E

at the centre on the difference of

the intercepted arcs.

74. The locus of the middle points of chords

drawn from the same point in the circumference is a circumference.

Let A B be a chord and AC a diameter from A. Let D be the middle point of A B. Join D to the centre O, and also join BC, then DO is parallel to BC (Example 43); hence A DO is a right angle.

The locus of D is therefore the locus of vertices of right-angled triangles on AO as base, and is therefore a circumference of which AO is diameter.

75. A B C is a triangle inscribed in a circle. Show that the straight line which bisects A and the straight line drawn from the centre perpendicular to BC meet on the circumference.

Let the perpendicular bisector of B C meet the

E

bisector of the angle A in D, and. when continued in the opposite direction let it meet the circumference in E. Draw the lines BE,. BCE, and suppose B E to cut A D

in F.

Now BAC=/ BEC; ... BAD BED.

=

Again in the triangles FED and FAB, 4 EFD LAF B.

Hence the third angles E DA, E B A are equal, and consequently D is on the circumference.

76. The circles described on the sides of any triangle as diameters will intersect in the sides or the sides produced of the triangle.

Let A B C be the triangle, and let AD be perpendicular from A to B C.

Then, since A B D is a right

angled triangle, the circle on

A B as diameter passes through B

D

D. Similarly, the circle on A C as diameter passes through D, and therefore intersects the former circle in D.

So also the circles on the diameters BA, B C intersect in the foot of the perpendicular from B on A C, and the circles on CA and C B intersect in the foot of the perpendicular from C on A B.

77. The feet of the perpendiculars drawn to the sides of a triangle from any point on the circumference which passes through its vertices lie in a straight line.