Main Concept

Let $\mathrm{AB}$ be the diameter of a circle, $\mathrm{AC}$ a side of an inscribed regular pentagon, $D$ the middle point of the arc $\mathrm{AC}$. Join $\mathrm{CD}$ and produce it to meet $\mathrm{BA}$ produced in $E$; join $\mathrm{AC}$, $\mathrm{DB}$ meeting in $F$, and draw $\mathrm{FM}$ perpendicular to $\mathrm{AB}$. Then, $\mathrm{EM}$ = ( $\mathrm{radius} \mathrm{of} \mathrm{circ}\mathrm{le}$ ).

At the beginning of the proof, you might need to note this fact:

The inscribed angle theorem states that an angle q inscribed in a circle is half of the central angle 2q, that subtends the same arc on the circle.

A consequence of this is: For four consecutive points $A\,B\, C\, and \mathrm{D}$ on a circle, $\angle \mathrm{ABC} \+ \angle \mathrm{ADC} \= \mathrm{\π}$.

Proof:

Let $O$ be the center of the circle and join $\mathrm{DA}$, $\mathrm{DM}$, $\mathrm{DO}$, $\mathrm{CB}$.

Now:                       $\angle \mathrm{ABC} \= \frac{\mathrm{\π}}{5}$,   

and,         $\angle \mathrm{ABD} \= \angle \mathrm{DBC} \= \frac{\mathrm{\π}}{10}$,

hence,                 $\angle \mathrm{AOD} \= \frac{\mathrm{\π}}{5} \.$

Further, the triangles $\mathrm{FCB}$, $\mathrm{FMB}$, are equal in all respects.

Therefore, in the triangles $\mathrm{DCB}$, $\mathrm{DMB}$, the sides $\mathrm{CB}$, $\mathrm{MB}$, being equal and $\mathrm{BD}$ common, while the angles $\mathrm{CBD}$, $\mathrm{MBD}$ are equal,

However,         $$\begin{gathered} \angle \mathrm{BCD} \+ \angle \mathrm{BAD} \= \mathrm{two} \mathrm{rectangles} \left(\= \mathrm{\π}\right) \\ \= \angle \mathrm{BAD} \+ \angle \mathrm{DAE} \\ \= \angle \mathrm{BMD} \+ \angle \mathrm{DMA}\, \end{gathered}$$

so that,                   $\angle \mathrm{DAE} \= \angle \mathrm{BCD}$,

and,                        $\angle \mathrm{BAD} \= \angle \mathrm{AMD}$.

Therefore,

                $\mathrm{AD} \= \mathrm{MD}$.

Now, in the triangle $\mathrm{DMO}$,

                             $\angle \mathrm{MOD} \= \frac{\mathrm{\π}}{5}$,

                             $\angle \mathrm{DMO} \= \frac{3 \mathrm{\π}}{5}$,

(because $\angle \mathrm{AMD}\=\angle \mathrm{MAD}\=\angle \mathrm{OAD}\=\angle \mathrm{ODA}\=\frac{2 \mathrm{\π}}{5}$).

Therefore,               $\angle \mathrm{ODM} \= \frac{\mathrm{\π}}{5} \= \angle \mathrm{MOD}$;

hence,                        $\mathrm{OM} \= \mathrm{MD}$.

Again:                      $$\begin{gathered} \angle \mathrm{EDA} \= \left(\mathrm{supplement} \mathrm{of} \mathrm{ADC}\right) \\ \= \angle \mathrm{CBA} \\ \= \frac{\mathrm{\π}}{5} \\ \= \angle \mathrm{ODM}\. \end{gathered}$$       

Therefore, in the triangles $\mathrm{EDA}$, $\mathrm{ODM}$,

                                           $\angle \mathrm{EDA} \= \angle \mathrm{ODM}$,

                                           $\angle \mathrm{EAD} \= \angle \mathrm{OMD}$,

and the sides $\mathrm{AD}$, $\mathrm{MD}$ are equal.

Hence, the triangles are equal in all respects and,

                                                 $\mathrm{EA} \= \mathrm{MO}$,

Therefore,                                 $\mathrm{EM} \= \mathrm{AO}$.

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