Given a square $ABCD$, let $P$ be a point within it. Join $PA, PB, PC, PD$ in the given figure. What is the ratio of the areas of the red region ($\Delta PAB$ and $\Delta PCD$) and the green region ($\Delta PBC$ and $\Delta PDA$)?                                                                                                                                                                                                                                          [Page No. 143]

Square ABCD is given and P is any interior point.

The red region consists of: $\Delta\text{PAB} + \Delta\text{PCD}$

The green region consists of: $\Delta\text{PBC} + \Delta\text{PDA}$

In a square:

opposite sides are parallel and equal.

Triangles on opposite sides with the same altitude have equal combined areas.

Thus,

$$\text{Area}(\Delta\text{PAB}) + \text{Area}(\Delta\text{PCD}) = \frac{1}{2} \times \text{Area of square}$$

Similarly,

$$\text{Area}(\Delta\text{PBC}) + \text{Area}(\Delta\text{PDA}) = \frac{1}{2} \times \text{Area of square}$$

Therefore, Red region area = Green region area

Hence, the ratio is 1 : 1.