$D$ and $E$ are points on sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$. Prove that $DE \parallel BC$.

$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.

In the figure,$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F$,show that:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$

In the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$.

$XY$ is a line parallel to side $BC$ of a triangle $ABC$. If $BE || AC$ and $CF || AB$ meet $XY$ at $E$ and $F$ respectively,show that $\operatorname{ar}(ABE) = \operatorname{ar}(ACF)$.

In the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.