Question 11

We will show that Area($\Delta$BPQ) equals Area($\Delta$DBC). Then we will use the midpoint property.

Step 1 — Relate $\Delta$DBC to $\Delta$ABC

D is the midpoint of AB. So, the base BD is half of AB. Triangles $\Delta$DBC and $\Delta$ABC share the same height from C to AB. Their areas are proportional to their bases.

$\text{Area }(\Delta\text{DBC}) = \frac{\text{BD}}{\text{AB}} \times \text{Area }(\Delta\text{ABC})$

$= \frac{1}{2} \times \text{Area }(\Delta\text{ABC})$

$\boxed{\text{Area }(\Delta\text{DBC}) = \frac{1}{2} \text{Area }(\Delta\text{ABC})}$

Step 2 — Use parallel lines property

We are given that CQ is parallel to PD. Consider triangles $\Delta$PDQ and $\Delta$PDC. They share the same base PD. Their vertices Q and C lie on the line CQ. Since CQ is parallel to PD, these triangles lie between the same parallel lines. Therefore, their areas are equal.

$\text{Area }(\Delta\text{PDQ}) = \text{Area }(\Delta\text{PDC})$

Step 3 — Express Area($\Delta$BPQ)

From the diagram, point D lies on the line segment BQ. So, the area of $\Delta$BPQ can be split. It is the sum of the areas of $\Delta$BDP and $\Delta$DPQ.

$\text{Area }(\Delta\text{BPQ}) = \text{Area }(\Delta\text{BDP}) + \text{Area }(\Delta\text{DPQ})$

Now, substitute the result from Step 2 into this equation.

$\text{Area }(\Delta\text{BPQ}) = \text{Area }(\Delta\text{BDP}) + \text{Area }(\Delta\text{PDC})$

Step 4 — Relate Area($\Delta$BPQ) to Area($\Delta$DBC)

From the diagram, point P lies on the line segment BC. So, the area of $\Delta$DBC can be split. It is the sum of the areas of $\Delta$BDP and $\Delta$DPC.

$\text{Area }(\Delta\text{DBC}) = \text{Area }(\Delta\text{BDP}) + \text{Area }(\Delta\text{DPC})$

We found in Step 3 that Area($\Delta$BPQ) = Area($\Delta$BDP) + Area($\Delta$DPC). By comparing these two expressions, we see they are equal.

$\text{Area }(\Delta\text{BPQ}) = \text{Area }(\Delta\text{DBC})$

Step 5 — Final conclusion

From Step 1, we know that Area($\Delta$DBC) is half of Area($\Delta$ABC). From Step 4, we established that Area($\Delta$BPQ) is equal to Area($\Delta$DBC). Therefore, we can conclude the desired result.

$\text{Area }(\Delta\text{BPQ}) = \frac{1}{2} \text{Area }(\Delta\text{ABC})$

$\boxed{\text{Area }(\Delta\text{BPQ}) = \frac{1}{2} \text{Area }(\Delta\text{ABC})}$

Answer

The proof shows that Area($\Delta$BPQ) = $\frac{1}{2}$ Area($\Delta$ABC).