Problem 1711

1. **Identify the areas of quadrilaterals $ATOC$ and $BTOC$**:
- We need to show that the areas of quadrilaterals $ATOC$ and $BTOC$ are equal.

2. **Calculate the area of $BTOC$**:
- The area of quadrilateral $BTOC$ can be split into the areas of triangles $BTC$ and $COT$.
- The area of triangle $BTC$ is given by:
$$A_{BTC} = \frac{1}{2} \cdot BT \cdot BC \cdot \sin \angle BTC$$
Since $\angle BTC = \angle B$, we have:
$$A_{BTC} = \frac{1}{2} \cdot BT \cdot BC \cdot \sin \angle B$$
- The area of triangle $COT$ is given by:
$$A_{COT} = \frac{1}{2} \cdot CT \cdot OC \cdot \sin \angle OCT$$
Since $\angle OCT = 90^\circ - \angle B$, we have:
$$A_{COT} = \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$$

3. Combine the areas:
- The total area of $BTOC$ is:
$$A_{BTOC} = A_{BTC} + A_{COT} = \frac{1}{2} \cdot BT \cdot BC \cdot \sin \angle B + \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$$

4. **Calculate the area of $ATOC$**:
- The area of quadrilateral $ATOC$ can be split into the areas of triangles $ATC$ and $COT$.
- The area of triangle $ATC$ is given by:
$$A_{ATC} = \frac{1}{2} \cdot AT \cdot AC \cdot \sin \angle ATC$$
Since $\angle ATC = \angle A$, we have:
$$A_{ATC} = \frac{1}{2} \cdot AT \cdot AC \cdot \sin \angle A$$
- The area of triangle $COT$ is the same as calculated before:
$$A_{COT} = \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$$

5. Combine the areas:
- The total area of $ATOC$ is:
$$A_{ATOC} = A_{ATC} + A_{COT} = \frac{1}{2} \cdot AT \cdot AC \cdot \sin \angle A + \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$$

6. Compare the areas:
- To show that $A_{ATOC} = A_{BTOC}$, we need to show:
$$\frac{1}{2} \cdot AT \cdot AC \cdot \sin \angle A + \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B = \frac{1}{2} \cdot BT \cdot BC \cdot \sin \angle B + \frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$$
- Since the term $\frac{1}{2} \cdot CT \cdot OC \cdot \cos \angle B$ is common on both sides, we need to show:
$$\frac{1}{2} \cdot AT \cdot AC \cdot \sin \angle A = \frac{1}{2} \cdot BT \cdot BC \cdot \sin \angle B$$

7. **Use the fact that $AT = BT$**:
- Since $T$ is the foot of the altitude from $C$, $AT = BT$.
- Therefore, we need to show:
$$AC \cdot \sin \angle A = BC \cdot \sin \angle B$$

8. Use the Law of Sines:
- By the Law of Sines in $\triangle ABC$:
$$\frac{AC}{\sin \angle B} = \frac{BC}{\sin \angle A}$$
- Rearranging, we get:
$$AC \cdot \sin \angle A = BC \cdot \sin \angle B$$

9. Conclusion:
- Since $AC \cdot \sin \angle A = BC \cdot \sin \angle B$, we have:
$$A_{ATOC} = A_{BTOC}$$
- Therefore, the areas of quadrilaterals $ATOC$ and $BTOC$ are equal.

$\blacksquare$