Bisecting angles of triangles within circles.

[Kassie]

Could someone help me with this puzzle?

Two distinct fixed points, A and B, are given on the circuference of circle C. A point P moves everywhere on C except that P does not equal A or B.

Prove that the bisecor of the angle APB subtended by the chord AB at the circumference passes through either one of two fixed points on C. Furthermore, prove that these two points are diametrically opposite each other.

Thanks,
Kassie.

[Christopher Henrich]

In article
<24800139.1213288172...@nitrogen.mathforum.org>,
Kassie <kestrel....@hotmail.co.uk> wrote:

The angle APB is equal to half the arc from A to B opposite the point P.

This fact will give you a hint as to where the bisector of APB meets the
arc.

--
Christopher J. Henrich
chen...@monmouth.com
htp://www.mathinteract.com

[Narasimham]

On Jun 12, 9:35 pm, Kassie <kestrel.starl...@hotmail.co.uk> wrote:

> Two distinct fixed points, A and B, are given on the circuference of circle C. A point P moves everywhere on C except that P does not equal A or B.
>
> Prove that the bisecor of the angle APB subtended by the chord AB at the circumference passes through either one of two fixed points on C. Furthermore, prove that these two points are diametrically opposite each other.

Let Q be a point on C in segment opposite to that of P. Consider the
following angles.

APQ = QPB ; 2 APB = 2 QPB; Angles subtended at center AOQ = QOB ; so,
Q is the center of arc AB. HTH

Narasimham