Problem 1471

1. Given Information and Setup:
- We are given a convex quadrilateral $ABCD$ with $\angle A + \angle D = 150^\circ$.
- We are also given that $\angle B < 150^\circ$ and $\angle C < 150^\circ$.
- We need to prove that the area of $ABCD$ is greater than $\frac{1}{4}(AB \cdot CD + AB \cdot BC + BC \cdot CD)$.

2. Intersection and Angle Calculation:
- Let $AB$ and $CD$ intersect at point $E$.
- Since $\angle A + \angle D = 150^\circ$, the external angle at $E$ is $180^\circ - 150^\circ = 30^\circ$. Thus, $\angle E = 30^\circ$.

3. Altitude Construction:
- Let $DF$ and $BG$ be the altitudes from $D$ and $B$ to $AB$ and $CD$ respectively.
- Since $\angle EBC > 30^\circ$ and $\angle ECB > 30^\circ$, it follows that $BC < EB$ and $BC < EC$.

4. Area Calculation:
- The area of $ABCD$ can be split into two triangles: $\triangle ADB$ and $\triangle DBC$.
- The area of $\triangle ADB$ is $\frac{1}{2} \cdot AB \cdot DF$.
- The area of $\triangle DBC$ is $\frac{1}{2} \cdot CD \cdot BG$.

5. Expressing Altitudes:
- Given $DF = \frac{1}{2} DE$ and $BG = \frac{1}{2} BE$, we can rewrite the areas:
$$\text{Area of } \triangle ADB = \frac{1}{2} \cdot AB \cdot \frac{1}{2} DE = \frac{1}{4} AB \cdot DE$$
$$\text{Area of } \triangle DBC = \frac{1}{2} \cdot CD \cdot \frac{1}{2} BE = \frac{1}{4} CD \cdot BE$$

6. **Total Area of $ABCD$:**
- The total area of $ABCD$ is:
$$\text{Area of } ABCD = \text{Area of } \triangle ADB + \text{Area of } \triangle DBC = \frac{1}{4} AB \cdot DE + \frac{1}{4} CD \cdot BE$$

7. Comparison with Given Expression:
- We need to show:
$$\frac{1}{4} AB \cdot DE + \frac{1}{4} CD \cdot BE > \frac{1}{4} (AB \cdot CD + AB \cdot BC + BC \cdot CD)$$
- Since $BC < EB$ and $BC < EC$, it follows that:
$$AB \cdot DE + CD \cdot BE > AB \cdot CD + AB \cdot BC + BC \cdot CD$$

8. Conclusion:
- Therefore, the area of $ABCD$ is greater than $\frac{1}{4}(AB \cdot CD + AB \cdot BC + BC \cdot CD)$.

$\blacksquare$