I is true. In $\Delta CDE$,
$\begin{array}{rcll}
\angle CDE & = & \dfrac{(5-2) \times 180^\circ}{5} & \text{($\angle$ sum of polygon)} \\
\angle CDE & = & 108^\circ
\end{array}$
Also,
$\begin{array}{rcll}
CD & = & ED & \text{(regular pentagon)} \\
\angle DCE & = & \angle DEC & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle DCE & = & \dfrac{1}{2} \times (180^\circ – \angle CDE) & \text{($\angle$ sum of $\Delta$)} \\
\angle DCE & = & \dfrac{1}{2} \times (180^\circ – 108^\circ) \\
\angle DCE & = & 36^\circ
\end{array}$
Consider $\Delta CDE$ and $\Delta DEA$.
$\begin{array}{rcll}
CD & = & DE & \text{(regular pentagon)} \\
DE & = & EA & \text{(regular pentagon)} \\
\angle CDE & = & \angle DEA & \text{(regular pentagon)}
\end{array}$
Therefore, $\Delta CDE \cong \Delta DEA \ \text{(S.A.S.)}$.
Hence, $\angle ADE = \angle ECD = 36^\circ\ \text{(corr. $\angle$s, $\cong \Delta$s)}$.
Consider $\Delta CDF$.
$\begin{array}{rcl}
\angle CDF & = & \angle CDE – \angle EDA \\
\angle CDF & =& 108^\circ – 36^\circ \\
\angle CDF & = & 72^\circ
\end{array}$
Also,
$\begin{array}{rcll}
\angle CFD & = & 180^\circ – \angle DCF – \angle CDF & \text{($\angle$ sum of $\Delta$)} \\
\angle CFD & = & 180^\circ – 36^\circ – 72^\circ \\
\angle CFD & = & 72^\circ
\end{array}$
Since $\angle CDF = \angle CFD$, then $CF = CD\ \text{(sides opp. eq. $\angle$s)}$.
II is true. In $\Delta AFD$ and $\Delta AFE$,
$\begin{array}{rcll}
CD & = & AE & \text{(regular pentagon)} \\
\angle DCF & = & \angle EAF & \text{(corr. $\angle$s, $\cong \Delta$s)} \\
\angle CFD & = & \angle AFE & \text{(vert. opp. $\angle$s)}
\end{array}$
Therefore, $\Delta AFD \cong \Delta AFE\ \text{(A.A.S.)}$.
Hence, $CF = AF\ \text{(corr. sides, $\cong \Delta$s)}$.
In $\Delta ABF$ and $\Delta CBF$,
$\begin{array}{rcll}
AB & = & CB & \text{(regular pentagon)} \\
BF & = & BF & \text{(common side)} \\
AF & = & CF & \text{(proved)} \\
\end{array}$
Therefore, $\Delta ABF \cong \Delta CBF\ \text{(S.S.S.)}$.
III is true. Since $\Delta ABF \cong \Delta CBF$, $\angle BFC = \angle BFA\ \text{corr. $\angle$s, $\cong \Delta$s)}$.
$\begin{array}{rcll}
\angle AFB & = & \dfrac{1}{2} ( 180^\circ – \angle AFE) & \text{(adj. $\angle$s on a st. line)} \\
\angle AFB & = & \dfrac{1}{2} (180^\circ – 72^\circ) \\
\angle AFB & = & 54^\circ
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\angle AFB + \angle EAF & = & 54^\circ + 36^\circ \\
& = & 90^\circ
\end{array}$
