$ABCD$ is a parallelogram. $P$ and $Q$ are any two points on side $AB$. What can you say about the ratio $\text{Area}(\Delta PCD) : \text{Area}(\Delta QCD)$?                                                                                                                                         [Page No. 142]

Both triangles $\Delta PCD$ and $\Delta QCD$ share the same base $CD$.

P and Q are both on line AB, which is parallel to CD.

Therefore, both triangles have the same perpendicular height (= height of parallelogram).

$$\text{Area}(\triangle PCD) = \frac{1}{2} \times CD \times h$$

$$\text{Area}(\triangle QCD) = \frac{1}{2} \times CD \times h$$

$$\therefore \text{Area}(\triangle PCD) : \text{Area}(\triangle QCD) = 1 : 1$$