ABCD is a square. E and F are respectively

Given in square $A B C D, E$ and $F$ are the mid-points of $B C$ and $C D$ respectively. Also, $R$ is the mid-point of $E F$.

To prove $\operatorname{ar}(A E R)=\operatorname{ar}(A F R)$.

Construction Draw $A N \perp E F$

Proof $\because \quad$ ar $(\triangle A E R)=\frac{1}{2} \times$ Base $\times$ Height

$=\frac{1}{2} \times E R \times A N$

$=\frac{1}{2} \times F R \times A N$

$[\because R$ is the mid-point of $E F$, so $E R=F R]$

$=\operatorname{ar}(\Delta \mathrm{AFR})$

Hence proved.