Question

N Euclidean geometry, let AB and CD be intersecting chords of a circle, where

A, C, B, and D lie on the circle in the given order. Let P be the point of
intersection. Suppose that the angle bisector of ÐAPD passes through the center
of the circle. Show that AB = CD.

Answer 1

To demonstrate that AB = CD, the properties of inscribed angles and symmetrical division by the center are used. By showing that the bisector of ÁAPD creates equal angles that subtend the same arcs, we conclude that the chords AB and CD are equal because they subtend equal arcs.

Step-by-step explanation:

To show that the chords AB and CD of a circle are equal when the angle bisector of ÁAPD passes through the center of the circle, we can use the properties of inscribed angles and the symmetry of the angle bisector. The segments AP and PD are bisected by the angle bisector and the center of the circle, which means that angles ÁAPB and ÁDPB are equal because they subtend the same arc PB. Similarly, angles ÁAPC and ÁDPC are equal because they subtend the same arc PC.

As the angle bisector of ÁAPD passes through the center, it bisects the angle into two equal parts, which means that ÁAPB + ÁBPB = ÁDPC + ÁCPB. Since ÁBPB and ÁCPB are angles on the same chord PB, they are equal. Thus, ÁAPB = ÁDPC.

Using the fact that angles ÁBPC and ÁAPD are vertical angles and therefore equal, and given that ÁAPB = ÁDPC, it follows that the remaining angles ÁBPA and ÁCPD are also equal. By the converse of the Inscribed Angle Theorem, this implies that AB and CD subtend arcs of equal measure, therefore they must be equal in length. Hence, AB = CD as required.