91Ó°ÊÓ

As$ABCDE$ is a regular pentagon and $DEFG$is a square then, and all the interior angles of a regular pentagon is 108 degrees then, role="math" localid="1648817128078" $AB=BC=CD=DE=AE=EF=FG=DG$.

All angles of a square are 90 degrees then $∠ABC=∠BCD=∠CDE=∠DEA=∠EAB=108°$, $∠DEF=∠EFG=∠FGD=∠GDA=90°$.

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

Then,

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

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

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

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

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

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

Then by an angle sum theorem,

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

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

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

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