$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:

In the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$

$ABCD$ is a square. If $AC = 16 \, cm$,then find the area of $ABCD$ in $cm^2$.

State whether each of the following statements is true or false:
$(1)$ Area of a parallelogram $= \text{base} \times \text{corresponding altitude}$.
$(2)$ Area of a rhombus $= \frac{1}{2} \times \text{Product of its diagonals}$.
$(3)$ Area of a square $= (\text{Side})^2$.

In $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.

In rhombus $ABCD$,$AC = 12 \, cm$ and $BD = 15 \, cm$,then $\operatorname{ar}(ABCD) = \dots \, cm^2$.