$ABCD$ is a square. $E$ and $F$ are respectively the midpoints of $BC$ and $CD$. If $R$ is the midpoint of $EF$,prove that $\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.

(N/A) $ABCD$ is a square. $E$ and $F$ are respectively the mid-points of $BC$ and $CD$. If $R$ is the midpoint of $EF$,we have to prove that $\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.
In $\triangle ABE$ and $\triangle ADF$,we have:
$AB = AD$ [Sides of a square are equal]
$\angle ABE = \angle ADF = 90^{\circ}$
$BE = DF$ [Since $E$ is the mid-point of $BC$ and $F$ is the mid-point of $CD$,and $BC = CD$]
Therefore,$\triangle ABE \cong \triangle ADF$ [By $SAS$ congruence rule]
This implies $AE = AF$ [$CPCT$] ... $(1)$
Now,in $\triangle AER$ and $\triangle AFR$,we have:
$AE = AF$ [From $(1)$]
$ER = RF$ [Given that $R$ is the midpoint of $EF$]
$AR = AR$ [Common side]
Therefore,$\triangle AER \cong \triangle AFR$ [By $SSS$ congruence rule]
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.