43.3. In a circle, mutually perpendicular diameters $A B$ and $C D$ are drawn, and an arbitrary point $M$ is taken on the arc $A C$. Find $M B + M D$, if $M A + M C = a$.

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43.3. Let the center of the circle be denoted by \( O \) and set \( AM = x \), \( CM = y \). Rotate triangle \( AMB \) around point \( O \) by an angle of \( 90^\circ \) so that point \( A \) moves to point \( C \) (see Fig. 31). Since angle \( AMB \) is a right angle, point \( M \) will move to some point \( E \) on the given circle such that \( CE \parallel MB \) (after a \( 90^\circ \) rotation, segment \( AM \), which is perpendicular to \( MB \), will move to segment \( CE \), which is parallel to \( MB \)). Therefore, trapezoid \( MBEC \) is isosceles, i.e., \( \angle MBE = \angle BMC \). But angle \( BMC \) is an inscribed angle. It subtends arc \( BC \), which contains \( 90^\circ \), i.e., \( \angle BMC = \angle MBE = 45^\circ \).

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-45.jpg?height=617&width=646&top_left_y=237&top_left_x=705)

Fig. 31

Drop perpendiculars from points \( C \) and \( E \) to line \( MB \), denoted by \( CK \) and \( EP \). Now it is easy to find that \( MB = MK + KP + PB = y / \sqrt{2} + x + y / \sqrt{2} = y \sqrt{2} + x \). Similarly, by rotating triangle \( CMD \) around point \( O \) so that point \( C \) moves to point \( A \), we find that \( MD = x \sqrt{2} + y \). Finally, we have: \( MB + MD = y \sqrt{2} + x + x \sqrt{2} + y = (x + y)(1 + \sqrt{2}) = a(1 + \sqrt{2}) \).