To determine the ratio \(\frac{AD}{BE}\), we examine the properties of triangle \(ABC\) and its altitudes.
Given:
- \(\angle BAC = 45\degree\)
- \(\angle ABC = \theta\)
As triangle \(ABC\) is right-angled and \(AD\) and \(BE\) are its altitudes, trigonometric ratios are employed to establish their relationship. The altitudes partition the triangle into smaller triangles:
- In triangle \(ABD\), \(\angle BAD = 45\degree\)
- In triangle \(ABE\), \(\angle ABE = \theta\)
Applying the sine definition in these right triangles yields:
\(AD = AB \cdot \sin 45\degree = \frac{AB}{\sqrt{2}}\)
\(BE = AB \cdot \cos \theta\)
Consequently, the ratio is:
\[\frac{AD}{BE} = \frac{AB \cdot \frac{1}{\sqrt{2}}}{AB \cdot \cos \theta} = \frac{1}{\sqrt{2} \cdot \cos \theta}\]
Multiplying the numerator and denominator by \(\sqrt{2}\) results in:
\[\frac{\sqrt{2}}{2 \cdot \cos \theta} = \sqrt{2} \cdot \sin \theta\]
The conclusive result is:
\(\sqrt{2}\sin\theta\)