Answer:
To prove that the bisector of angle APQ passes through O, we can utilize the properties of angles and intersecting chords in a circle. Here's a step-by-step proof:
Step 1: Consider triangle AOB. Since circle L passes through O, we know that OA and OB are radii of circle L.
Step 2: Since P lies on arc AOB, angle AOP and angle BOP are inscribed angles of the circle K.
Step 3: By the Inscribed Angle Theorem, we know that the measure of angle AOP is equal to half the measure of arc AB, and the measure of angle BOP is also equal to half the measure of arc AB.
Step 4: Since angle AOP and angle BOP share the same arc AB, they have equal measures. Therefore, angle AOP = angle BOP.
Step 5: By the Vertical Angles Theorem, we know that angle AOP and angle QOP are vertical angles, and hence, they are congruent. Therefore, angle AOP = angle QOP.
Step 6: By transitivity, we can conclude that angle BOP = angle QOP.
Step 7: Since angle AOP = angle BOP and angle AOP = angle QOP, we can deduce that angle BOP = angle QOP.
Step 8: By the Converse of the Inscribed Angle Theorem, if two angles intercept the same arc AB on circle K and have equal measures, then the corresponding chords that connect the endpoints of the intercepted arc to the vertices of the angles are equal in length.
Step 9: Applying the Converse of the Inscribed Angle Theorem, we can conclude that chord BP is congruent to chord BQ.
Step 10: Since O is the center of circle K, chord BP is a radius of circle K. By the definition of a radius, it passes through the center O.
Step 11: Since chord BP is congruent to chord BQ, and BP passes through O, we can infer that chord BQ also passes through O.
Step 12: Finally, since angle APQ is formed by rays AQ and QP, the bisector of angle APQ will pass through the point of intersection of chords BP and BQ, which is the center O.
Therefore, the bisector of angle APQ passes through O, as proven.