Two chords AB and CD of a circle intersect internally at a point P. If angle APC = 50°, and the minor arc AC subtends a central angle of 60°, find the central angle subtended by the minor arc BD.

Solution

Solution: When two chords intersect inside a circle, \(\angle APC=\frac12(\text{arc AC}+\text{arc BD})\). Substituting, \(50^\circ=\frac12(60^\circ+\text{arc BD})\). Therefore, \(100^\circ=60^\circ+\text{arc BD}\), giving the central angle for minor arc BD as 40°. Trick: The angle formed by two intersecting chords inside a circle is half the sum of the intercepted arcs.