If a triangle and a parallelogram are on the same base and between the same parallels,then the ratio of the area of the triangle to the area of the parallelogram is:

In the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.

In square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.

$(1)$ In $\Delta ABC$,$AD$ is an altitude. If $BC = 8 \text{ cm}$ and $AD = 5 \text{ cm}$,then $\text{ar}(\Delta ABC) = \dots \text{ cm}^2$.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.

If the mid-points of the sides of a quadrilateral are joined in order,prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral.

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.