Problem 1710

1. Express the cotangent sums in terms of sines and cosines:

$$\cot \angle BAC + \cot \angle DAC = \frac{\cos \angle BAC}{\sin \angle BAC} + \frac{\cos \angle DAC}{\sin \angle DAC}$$

Using the sum-to-product identities, we can rewrite this as:

$$\cot \angle BAC + \cot \angle DAC = \frac{\sin (\angle BAC + \angle DAC)}{\sin \angle BAC \cdot \sin \angle DAC}$$

Since $\angle BAC + \angle DAC = \angle DAB$, we have:

$$\cot \angle BAC + \cot \angle DAC = \frac{\sin \angle DAB}{\sin \angle BAC \cdot \sin \angle DAC}$$

2. Similarly, express the other cotangent sum:

$$\cot \angle BCA + \cot \angle DCA = \frac{\cos \angle BCA}{\sin \angle BCA} + \frac{\cos \angle DCA}{\sin \angle DCA}$$

Using the sum-to-product identities, we can rewrite this as:

$$\cot \angle BCA + \cot \angle DCA = \frac{\sin (\angle BCA + \angle DCA)}{\sin \angle BCA \cdot \sin \angle DCA}$$

Since $\angle BCA + \angle DCA = \angle BCD$, we have:

$$\cot \angle BCA + \cot \angle DCA = \frac{\sin \angle BCD}{\sin \angle BCA \cdot \sin \angle DCA}$$

3. Form the ratio of the two cotangent sums:

$$\frac{\cot \angle BAC + \cot \angle DAC}{\cot \angle BCA + \cot \angle DCA} = \frac{\frac{\sin \angle DAB}{\sin \angle BAC \cdot \sin \angle DAC}}{\frac{\sin \angle BCD}{\sin \angle BCA \cdot \sin \angle DCA}}$$

Simplifying this, we get:

$$\frac{\cot \angle BAC + \cot \angle DAC}{\cot \angle BCA + \cot \angle DCA} = \frac{\sin \angle DAB \cdot \sin \angle BCA \cdot \sin \angle DCA}{\sin \angle BCD \cdot \sin \angle BAC \cdot \sin \angle DAC}$$

4. Relate the angles to the sides of the triangles:

Using the Law of Sines in triangles $\triangle DAB$ and $\triangle BCD$:

$$\frac{\sin \angle DAB}{\sin \angle BCD} = \frac{AB \cdot AD}{BC \cdot CD}$$

Therefore, we can write:

$$\frac{\cot \angle BAC + \cot \angle DAC}{\cot \angle BCA + \cot \angle DCA} = \frac{AB \cdot AD \cdot \sin \angle DAB}{BC \cdot CD \cdot \sin \angle BCD}$$

5. Relate the areas of the triangles:

The area of $\triangle DAB$ is given by:

$$|\triangle DAB| = \frac{1}{2} AB \cdot AD \cdot \sin \angle DAB$$

Similarly, the area of $\triangle BCD$ is given by:

$$|\triangle BCD| = \frac{1}{2} BC \cdot CD \cdot \sin \angle BCD$$

Therefore, we have:

$$\frac{|\triangle DAB|}{|\triangle BCD|} = \frac{AB \cdot AD \cdot \sin \angle DAB}{BC \cdot CD \cdot \sin \angle BCD}$$

6. **Relate the areas to the segments $AP$ and $PC$:**

Since $P$ is the intersection of the diagonals $AC$ and $BD$, the ratio of the areas of $\triangle DAB$ and $\triangle BCD$ is equal to the ratio of the segments $AP$ and $PC$:

$$\frac{|\triangle DAB|}{|\triangle BCD|} = \frac{AP}{PC}$$

Therefore, we have:

$$\frac{\cot \angle BAC + \cot \angle DAC}{\cot \angle BCA + \cot \angle DCA} = \frac{AP}{PC}$$

$\blacksquare$