91Ó°ÊÓ

As$ABCDE$is a regular pentagon and$DEFG$is a square then,$AB=BC=CD=DE=AE=EF=FG=DG$and all the interior angles of a regular pentagon is 108 degrees then, $∠ABC=∠BCD=∠CDE=∠DEA=∠EAB=108°$.

All angles of a square are 90 degrees then, localid="1649848898958" $∠DEF=∠EFG=∠FGD=∠GDA=90°$.

Consider$\mathrm{Δ}AEF$ in which $AE=EF$, therefore$\mathrm{Δ}AEF$ is an isosceles triangle, then $m∠EAF=m∠EFA$.

Then,

$$\begin{gathered} m∠AEF+m∠AED+m∠FED=360 \\ m∠AEF+108+90=360 \\ m∠AEF=360−198 \\ =162 \end{gathered}$$

Consider$\mathrm{Δ}AEF$ then by an angle sum theorem,

$$\begin{gathered} m∠AEF+m∠EFA+m∠EAF=180 \\ 162+2m∠EAF=180 \\ 2m∠EAF=18 \\ m∠EAF=9° \end{gathered}$$

$∠AFD$can be written as$∠AFD=∠EFA+EFD$ then

$$\begin{gathered} m∠AFD=m∠EFA+m∠EFD \\ =9+45 \\ =54° \end{gathered}$$

Consider$\mathrm{Δ}AED$ in which $AE=ED$, therefore$\text{Δ}AED$ is an isosceles triangle, then $m∠DAE=m∠EDA$.

Then by an angle sum theorem,

$$\begin{gathered} m∠DAE+m∠EDA+m∠AED=180 \\ 2m∠DAE+108=180 \\ 2m∠DAE=72 \\ m∠DAE=36° \end{gathered}$$

$∠DAF$can be written as $∠DAF=∠DAE+∠EAF$then

$$\begin{gathered} m∠DAF=m∠DAE+m∠EAF \\ =36+9 \\ =45° \end{gathered}$$

Therefore, the measures of$∠EAF,∠AFD$ and$∠DAF$ is$9°,54°$and$45°$ respectively.