In the adjacent figure, ABCD is a rectangle.

Refer to the figure given in the book.

In $\Delta B M C$ and $\Delta D N A$ :

$\angle D N A=\angle B M C=90^{\circ}$

$\angle B C M=\angle D A N \quad$ (alternate angles)

$B C=D A \quad$ (opposite sides)

By AAS congruency criteria :

$\Delta B M C \cong \Delta D N A$

Yes, it is true that $B M$ is equal to $D N$. (by corresponding parts of congruent triangles $B M C$ and $D N A$ )