+10 In a right triangle the median has the same length as half the hypotenuse because of the circumcircle of the triangle. Therefore:
\(\triangle{ZCA}\) is isoceles and \(\angle{ZAC}=\angle{ZCA}=90-\angle{ABX}\)
\(\angle{ABX}=90-\angle{BAX}=90-(\frac{\angle{BAC}}{2}-\angle{XAY})=90-(45-22)=67\)
Thus \(\angle{ZAC}=\angle{ZCA}=90-67=\boxed{23}\)
+10 In a right triangle the median has the same length as half the hypotenuse because of the circumcircle of the triangle. Therefore:
\(\triangle{ZCA}\) is isoceles and \(\angle{ZAC}=\angle{ZCA}=90-\angle{ABX}\)
\(\angle{ABX}=90-\angle{BAX}=90-(\frac{\angle{BAC}}{2}-\angle{XAY})=90-(45-22)=67\)
Thus \(\angle{ZAC}=\angle{ZCA}=90-67=\boxed{23}\)
