MATHCOUNTS Exercises – 3

Solution 1 $$ \begin{align} & \because m\angle{ABE}=\dfrac{1}{2}\cdot m\overparen{AE\ } \\ & \therefore m\overparen{AE\ }=2\cdot m\angle{ABE}=2\cdot 35^\circ=70^\circ \\ & \because m\angle{AXE}=\dfrac{m\overparen{AE\ }-m\overparen{CD\ }}{2} \\ & \therefore m\overparen{CD\ }=m\overparen{AE\ }-2\cdot m\angle{AXE}=70^\circ-2\cdot 15^\circ=40^\circ \blacksquare \\ \end{align} $$

Solution 2 Draw line \(\overline{OD}\), \(\overline{OE}\) and chord \(\overline{BF}\) passing thru point O.

\(\because \triangle{AOB}\) is isosceles
\(\therefore m\angle{AOF}=m\angle{OBA}+m\angle{OAB}=2\cdot m\angle{OBA}\)
\(\because \triangle{BOE}\) is isosceles
\(\therefore m\angle{EOF}=m\angle{OBE}+m\angle{OEB}=2\cdot m\angle{OBE}\)
\(\therefore m\angle{AOE}=m\angle{AOF}-m\angle{EOF}=2\cdot m\angle{OBA}-2\cdot m\angle{OBE}\)
\(\ \ \ \ \ \ =2\cdot(m\angle{OBA}-m\angle{OBE})=2\cdot m\angle{ABE}=2\cdot 35^\circ=70^\circ\)
\(\therefore m\angle{OED}=m\angle{AOE}-m\angle{AXE}=70^\circ-15^\circ=55^\circ\)
\(\because \triangle{DOE}\) is isosceles
\(\therefore m\angle{DOE}=180^\circ-2\cdot m\angle{OED}=180^\circ-2\cdot55^\circ=70^\circ\)
\(\therefore m\overparen{CD\ }=m\angle{COD}=180^\circ-m\angle{DOE}-m\angle{AOE}=180^\circ-70^\circ-70^\circ=40^\circ\)
\(\blacksquare\)