1. The problem states that \(\overline{AD}\) and \(\overline{CE}\) are diameters of circle \(P\). 2. Since \(AD\) and \(CE\) are diameters, \(A\), \(D\), \(C\), and \(E\) lie on the circle with center \(P\). 3. The angle between radius \(PB\) and \(PC\) is given as \(38^\circ\). 4. The angle between radius \(PD\) and \(PE\) is given as \(93^\circ\). 5. Since \(AD\) is a diameter, points \(A\) and \(D\) lie on a straight line through center \(P\), forming a 180-degree arc. 6. Similarly, \(CE\) is also a diameter formed by points \(C\) and \(E\). 7. The key is to find the measure of minor arc \(\widehat{AB}\) which lies between points \(A\) and \(B\). 8. To find arc \(\widehat{AB}\), note that it is intercepted by central angle \(\angle APB\). 9. Since \(\angle BPC = 38^\circ\) and \(C\), \(E\) are opposite ends of a diameter, the angle \(APB\) is supplementary to \(BPC + CPD\) or related to these angles. 10. The arc between points can be calculated based on the relationships that diameters create two semicircles of 180 degrees. However, given the complexity, reconsider the approach: - Diameter \(AD\) divides the circle into two semicircles: arcs \(\widehat{AB} + \widehat{BC}\) equal 180 degrees. - \(\angle BPC = 38^\circ\) is the central angle subtending arc \(\widehat{BC}\). - Therefore, arc \(\widehat{BC} = 38^\circ\). - Hence, arc \(\widehat{AB} = 180^\circ - 38^\circ = 142^\circ\). Final answer: The arc measure of minor arc \(\widehat{AB}\) is \(142^\circ\).