Problem 1718

1. Assign Coordinates:
We start by assigning coordinates to the vertices of the tetrahedron. Let:
$$A = (0, 0, 0), \quad B = (b, 0, 0), \quad C = (0, c, 0), \quad D = (0, 0, 1)$$
This ensures that the angles $\angle BAC$, $\angle CAD$, and $\angle DAB$ are all $90^\circ$.

2. Vectors Calculation:
Calculate the vectors $\overrightarrow{DB}$ and $\overrightarrow{DC}$:
$$\overrightarrow{DB} = (b, 0, -1), \quad \overrightarrow{DC} = (0, c, -1)$$

3. Dot Product and Magnitudes:
Compute the dot product $\overrightarrow{DB} \cdot \overrightarrow{DC}$:
$$\overrightarrow{DB} \cdot \overrightarrow{DC} = b \cdot 0 + 0 \cdot c + (-1) \cdot (-1) = 1$$
Compute the magnitudes $|\overrightarrow{DB}|$ and $|\overrightarrow{DC}|$:
$$|\overrightarrow{DB}| = \sqrt{b^2 + 0^2 + (-1)^2} = \sqrt{b^2 + 1}$$
$$|\overrightarrow{DC}| = \sqrt{0^2 + c^2 + (-1)^2} = \sqrt{c^2 + 1}$$

4. **Cosine of Angle $\angle BCD$:**
Using the dot product formula, we find $\cos \angle BCD$:
$$\cos \angle BCD = \frac{\overrightarrow{DB} \cdot \overrightarrow{DC}}{|\overrightarrow{DB}| \cdot |\overrightarrow{DC}|} = \frac{1}{\sqrt{b^2 + 1} \cdot \sqrt{c^2 + 1}}$$
Therefore,
$$\cos^2 \angle BCD = \left( \frac{1}{\sqrt{b^2 + 1} \cdot \sqrt{c^2 + 1}} \right)^2 = \frac{1}{(b^2 + 1)(c^2 + 1)}$$

5. **Sine of Angle $\angle BCD$:**
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we get:
$$\sin^2 \angle BCD = 1 - \cos^2 \angle BCD = 1 - \frac{1}{(b^2 + 1)(c^2 + 1)}$$
Simplify the expression:
$$\sin^2 \angle BCD = \frac{(b^2 + 1)(c^2 + 1) - 1}{(b^2 + 1)(c^2 + 1)} = \frac{b^2 + c^2 + b^2c^2}{(b^2 + 1)(c^2 + 1)}$$

6. **Area of Triangle $BCD$:**
The area of triangle $BCD$ is given by:
$$\text{Area}(BCD) = \frac{1}{2} |\overrightarrow{DB}| |\overrightarrow{DC}| \sin \angle BCD$$
Squaring both sides, we get:
$$\text{Area}(BCD)^2 = \left( \frac{1}{2} \right)^2 |\overrightarrow{DB}|^2 |\overrightarrow{DC}|^2 \sin^2 \angle BCD$$
Substitute the values:
$$\text{Area}(BCD)^2 = \frac{1}{4} (b^2 + 1)(c^2 + 1) \cdot \frac{b^2 + c^2 + b^2c^2}{(b^2 + 1)(c^2 + 1)} = \frac{1}{4} (b^2 + c^2 + b^2c^2)$$

7. **Areas of Triangles $BAC$, $CAD$, and $DAB$:**
The areas of triangles $BAC$, $CAD$, and $DAB$ are:
$$\text{Area}(BAC) = \frac{1}{2} bc, \quad \text{Area}(CAD) = \frac{1}{2} c, \quad \text{Area}(DAB) = \frac{1}{2} b$$
Squaring these areas, we get:
$$\text{Area}(BAC)^2 = \left( \frac{1}{2} bc \right)^2 = \frac{1}{4} b^2 c^2$$
$$\text{Area}(CAD)^2 = \left( \frac{1}{2} c \right)^2 = \frac{1}{4} c^2$$
$$\text{Area}(DAB)^2 = \left( \frac{1}{2} b \right)^2 = \frac{1}{4} b^2$$

8. Summing the Areas:
Summing the squares of the areas of triangles $BAC$, $CAD$, and $DAB$, we get:
$$\text{Area}(BAC)^2 + \text{Area}(CAD)^2 + \text{Area}(DAB)^2 = \frac{1}{4} b^2 c^2 + \frac{1}{4} c^2 + \frac{1}{4} b^2 = \frac{1}{4} (b^2 + c^2 + b^2c^2)$$

9. Conclusion:
We have shown that:
$$\text{Area}(BAC)^2 + \text{Area}(CAD)^2 + \text{Area}(DAB)^2 = \text{Area}(BCD)^2$$
Therefore, the given equation is satisfied.

$\blacksquare$