The medians AD and BE of the triangle with vertices `A(0,b),B(0,0)` and `C(a,0)` are mutually perpendicular. Prove that `a^2=2b^2`.

Explanation

To prove that \( a^2 = 2b^2 \) given that the medians \( AD \) and \( BE \) of triangle \( ABC \) are mutually perpendicular, we will follow these steps:

### Step 1: Find the coordinates of the midpoints D and E
- The coordinates of point \( A \) are \( (0, b) \), point \( B \) are \( (0, 0) \), and point \( C \) are \( (a, 0) \).
- The midpoint \( D \) of side \( BC \) can be calculated as:
\[ D = \left( \frac{0 + a}{2}, \frac{0 + 0}{2} \right) = \left( \frac{a}{2}, 0 \right) \]
- The midpoint \( E \) of side \( AC \) can be calculated as:
\[ E = \left( \frac{0 + a}{2}, \frac{b + 0}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right) \]

### Step 2: Calculate the slopes of the medians AD and BE
- The slope of median \( AD \) (from \( A \) to \( D \)):
\[ m_{AD} = \frac{y_D - y_A}{x_D - x_A} = \frac{0 - b}{\frac{a}{2} - 0} = \frac{-b}{\frac{a}{2}} = \frac{-2b}{a} \]
- The slope of median \( BE \) (from \( B \) to \( E \)):
\[ m_{BE} = \frac{y_E - y_B}{x_E - x_B} = \frac{\frac{b}{2} - 0}{\frac{a}{2} - 0} = \frac{\frac{b}{2}}{\frac{a}{2}} = \frac{b}{a} \]

### Step 3: Use the condition of perpendicularity
- Since the medians \( AD \) and \( BE \) are mutually perpendicular, the product of their slopes must equal \(-1\):
\[ m_{AD} \cdot m_{BE} = -1 \]
Substituting the slopes we found:
\[ \left( \frac{-2b}{a} \right) \cdot \left( \frac{b}{a} \right) = -1 \]
Simplifying this gives:
\[ \frac{-2b^2}{a^2} = -1 \]

### Step 4: Solve for \( a^2 \)
- Rearranging the equation:
\[ 2b^2 = a^2 \]
Therefore, we can conclude:
\[ a^2 = 2b^2 \]

### Conclusion
Thus, we have proved that \( a^2 = 2b^2 \).

---