MathFun4Kids

Hint Draw line $\overline{OC}$ and $\overline{PF}$

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
\(\therefore m\angle{DOE}=180^\circ-2\cdot m\angle{OED}=180^\circ-2\cdot55^\circ=70^\circ\)
\(\therefore m\overparen{CD\ }=m\angle{COD}=180^\circ-m\angle{DOE}-m\angle{AOE}=180^\circ-70^\circ-70^\circ=40^\circ\)
\(\blacksquare\)

Solution Draw lines $\overline{AB}$ and $\overline{AC}$ $$\begin{align} &\because major\ \overparen{BC\ }=230^\circ \\ &\therefore minor\ \overparen{BC\ }=360^\circ-major\ \overparen{BC\ }=360^\circ-230^\circ=130^\circ \\ &\therefore m\angle{BAC}=minor\ \overparen{BC\ }=130^\circ \\ &\because\ \overline{AB} = \overline{AC} \\ &\therefore \triangle{ABC}\ is\ isosceles \\ &\therefore m\angle{ABC}=m\angle{ACB}=\dfrac{180^\circ-m\angle{BAC}}{2}=\dfrac{180^\circ-130^\circ}{2}=25^\circ \\ &\because\ \overleftrightarrow{BD}\ is\ tangent\ to\ circle\ A\ at\ B \\ &\therefore\ \overline{AB} \perp \overleftrightarrow{BD} \\ &\therefore m\angle{ABD}=90^\circ \\ &\therefore m\angle{ABC}+m\angle{CBD}=m\angle{ABD}=90^\circ \\ &\therefore m\angle{CBD}=90^\circ-m\angle{ABC}=90^\circ-25^\circ=65^\circ \blacksquare \\ \end{align} $$

Solution 1 Draw line $\overline{AC}$. $$ \begin{align} &\because\ ABCDEFGHI\ \text{is a regular nonagon} \\ &\therefore \triangle{ABC}, \triangle{AIH}, \triangle{ACH} \text{are isosceles triangle} \\ &\therefore m\angle{HIA}=m\angle{IAB}=m\angle{ABC}=\dfrac{(9-2)\cdot 180^\circ}{9}=140^\circ\\ &\because \triangle{HIA} \cong \triangle{ABC} \\ &\therefore m\angle{IAH}=m\angle{BAC}=\dfrac{180^\circ-m\angle{ABC}}{2}=\dfrac{180^\circ-140^\circ}{2}=20^\circ \\ &\therefore m\angle{CAH}=m\angle{IAB}-m\angle{IAH}-m\angle{BAC}=140^\circ-20^\circ-20^\circ=100^\circ \\ &\therefore m\angle{AHC}=\dfrac{180^\circ-m\angle{CAH}}{2}=\dfrac{180^\circ-100^\circ}{2}=40^\circ \\ \end{align} $$

Solution 2 $$ \begin{align} &\because\ ABCDEFGHI\ \text{is a regular nonagon} \\ &\therefore m\overparen{AB}=m\overparen{BC}=\dfrac{180^\circ}{9}=20^\circ \\ &\therefore m\triangle{AMC}=m\overparen{AB}=m\overparen{BC}=20^\circ+20^\circ=40^\circ \\ \end{align} \\ $$